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- CHAPTER 03

CurrentElectricity

Chapter Analysis w.r.t: Lost 3 Yeor's Boord ExomsThe analysis given here gives you an analytical picture of this chapter and will helpyou to identifythe concepts of the chapter that are to befocussed morefrom exam point of view.

Number of Questions asked in last 3 years

Very Short Answer (1 mark) 1 Q 1 Q I 1 Q

20162015 2017

Short Type I Answer (2 m~u~s) 1 Q _ _1 <l. +__~_g__1-~9_-+-I__Short TYI:eII Answer (3 mark_sl__ ~_ _ ~9 -------r----------+---------J-------

valu~~:~:~::;t~::~:~~larksl --------- --------- 0-- ---------t--~--~-f---~

1----,---------- ---,-----------------+------,c----------IDelhi All India Delhi AllIndia Delhi AllIndia

• In 2015, in Delhi set, one question of 1mark based on J- V graph, one numerical questionof 2 marks based on Kirchhoff's rule and one question of 3 marks based on Drift Velocityand Relaxation Time were asked.'In Ail India set, one question of 1mark based on GraphType question on Ohm's Law, one question of 2 marks based on Kirchhoff's Law andBalance Condition of Wheatstone Bridge and one numerical question of 3 marks basedon Electric Circuit were asked.

• In 2016, only one numerical question 2 marks based on Circuit in All India set.• In 2017, in All India set, one question of 1mark based on Joule's Law of Heating and one

question of 5 marks based on Working of the Meter Bridge and neumerical based on itwere asked. In Delhi set, only one question of 5 marks based on Kirchhoff's Law wasasked.

On the basis of above analysis, it can be said that from exam point of view Kirchhoff's Rule, DriftVelocity, Relaxation Time and Meter Bridge are most important concepts of the chapter.

[TOPIC 1] Electric Conduction, Ohm's Law andResistance

1.1Electric CurrentThe directed rate of flow of electric chargethrough any cross-section of a conductor isknown as electric current.If !!Q charge flows in time !!t, then current at

. . I lim!!Q dQany time t IS = - = -!>J~ o!!t dt

Also, I = '1. = net t

[.: q = neJwhere, n = number of charged particlesconstitute the electric current.The direction of the current is in the flow ofpositive charge and opposite to the directionof flow of negative charge.SI unit of current is ampere and it isrepresented by A .

1 A = 1 cot..:.:=nb (C) = 10 s1 second (s)

NOTE Current is a scalar quantity.

Current DensityThe current density at a point in a conductoris the ratio of the current at that point in theconductor to the area of cross-section of theconductor at that point.If a current I is distributed uniformly overthe cross-section A of a conductor, then the

densi h .. J Icurrent ensity at t at pomt IS =-.A

NOTE Current density is a vector quantity.

Electric Current in Conductors(I) In Case of a Metallic ConductorAmong the solids, all metals are goodconductors of electricity. Free electrons arethe cause of conductance of electricity inthem.

(Ii) In Case of a Solid ConductorIn case of solid conductor (i.e. Cu, Fe, Ag, etc) atomsare tightly bound to each other but there are largenumber of free electrons in them to conduct.

(Iii) In Case of a Liquid ConductorIn case of a liquid conductor like electrolytic solution,there are positive and negative charged ions which canmove on applying electric field to conduct.

1.2 Ohm's LawAt constant temperature, the potential difference Vacross the ends of a given metallic wire (conductor) inan electric circuit is directly proportional to the currentflowing through it.

V cc I or V = IR

where, R = resistance of conductorThe variation of current w.r.t. applied potentialdifference is shown with the help of following graph.

I

"'-------vNo effect of V and I on R because as V increase, Iincrease but R remains the same.

Resistance of a ConductorIt is defined as the ratio of potential difference appliedacross the ends of the conductor to the current flowingthrough it.

Mathematically, R = ~I

SI unit is ohm (0).

Resistance can also be written as, R = p!:A

where, L = length of the conductor, A = area ofcross-section and p = constant. known as resistivity ofthe material. It depends upon the nature of thematerial and temperature of the conductor.

76

Effect of Temperature onResistanceTemperature coefficient of resistance averagedover the temperature range tl °C to t2°C is given by

R2 -RIex = -=----'--RI (t2 - tl)

where, RI and R2 are the resistances at tl °C andt2°C, respectively.

Drift VelocityIt is defined as the average velocity with whichthe free electrons move towards the positive endof a conductor under the influence of an externalelectric field applied across the conductor.

eEThus, vd = -'tm

where, 't = relaxation time, E = electric field,m = mass of the electron, e = electronic chargeThe drift velocity of electron is of the order oflO-4ms-l.

MobilityThe ratio of the drift velocity of the electrons tothe applied electric field is known as mobility. It• Vd q'tIS expressed as, Jl = - = -

E mIts SI unit is m2s-IV-I•

1.3 Resistivity of VariousMaterial

Resistivity of various material is defined as theresistance of unit cube of a material of the givenconductor

mp = ne2't

It depends upon the following factors

(i) p = ..!:.., where n = number of free electronn

(ii) p = ..!:.., where 't = average relaxation time of't

free electron.

IZl Chopterwise CBSE Solved Papers PHYSICS

Colour Code of Carbon ResistanceThe colour code of carbon resistor remains in theform of coaxial rings.The first band represents the first significant figure,second band represents second significant figureand third band represents multiplier (i.e. power often). The fourth band represents tolerance.

Black Brown Red Orange Yellow

I I I I IB B ROY

I I I I Io 2 3 4

Green Blue Violet Grey White

I I I I Iof Great Britain had Very Good Wife

I I I I I5 6 7 8 9

--f) )))) (j-Carbonresistor

Tolerance UnitGold 5%Silver 10%No colour 20%

1.4 Conductance andConductivity

ConductanceIt is defined as the reciprocal of resistance ofconductor.

1G=-R

Its SI unit is mho (Q-I) or siemen (S). Thedimensional formula of conductance is[M-IL-2T-3 A 2].

ConductivityIt is defined as the reciprocal of resistivity of aconductor. It is expressed as, o = ..!:..

p

CHAPTER 3 : Current Electricity

Its SI unit is mho per metre (n-l/ m ).

Relation between }, o and EThe relation between the current flowing throughthe conductor and drift velocity of electron isgiven by J = o EIt is microscopic form of Ohm's law.

SuperconductivityThe resistivity of certain metals or alloys drops tozero when they are cooled below a certaintemperature. This is called superconductivity. Itwas observed by Prof. Kamerlingh in 1911.

Some Important Units(i) Resistance - Ohm (n)(ii) Resistivity - Ohm-metre (n-m)

(iii) Conductance (i) -Mho or n-l orSiemen (S)

(iv) Current density - A/m2

NOTE If a conductor is stretched or compresses to n times oforiginal length, then

I'=nl ~ R'=n2R

where, R' = new resistance and R = original resistance.

1.5 Combinations ofResistors

There are two types of combinations of resistors

Series CombinationIn this combination, different resistors areconnected end to end.Equivalent resistance can be obtained by theformula,

Req = RI + R2 + ... + Rn

R, R2 R3 Rn~ .... -IWVV'-

NOTE The total resistance in the series combination is morethan the greatest resistance in the circuit.

77

Parallel CombinationIn this combination, first end of all the resistorsare connected to one point and last end of all theresistors are connected to other point.Equivalent resistance can be obtained by theformula,

R,

A?---------~~--------~B

Rn

NOTE The total resistance in parallel combination is less thanthe least resistance of the circuit.• If n identical resistors each of resistance rare

connected in

(i) series combination, Req = nr

(ii) parallel combination, Req = ~n

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions

1. J.V graph for a metallic wire at twodifferent temperatures 7;. and T2 is asshown in the figure below. Which of thetwo temperature is lower and why?Delhi 2015

v-

78

2. Graph showing the variation of currentversus voltage for a material GaAs isshown in the figure. Identify the region

(i) of negative resistance.(ii) where Ohm's law is obeyed.

All Indio 2015

1•....E~ B:::J()L...-=- _

Voltagev-3. Plot a graph showing the variation of

resistivity of a conductor withtemperature. Foreign 2015

4. How does the random motion of freeelectrons in a conductor get affected whena potential difference is applied across itsends? Delhi 2014

5. Plot a graph showing variation of currentversus voltage for the material GaAs.Deihl 2014

6. Show variation of resistivity of copperas a function of temperature in graph.Deihl 2014; All Indio 2014

7. Define the term drift velocity of chargecarriers in a conductor and write itsrelationship with the current flowingthrough it. Deihl 2014

8. Define the term electrical conductivity ofa metallic wire. Write its SI unit. Deihl 2014

9. Show variation of resistivity of Si withtemperature in graph. Deihl 2014, 12

10. Define the term mobility of chargecarriers in a conductor. Write its SI unit.Deihl 2014

11. How does one explain increase inresistivity of a metal with increase oftemperature? All Indio 2014C

12. Write a relation between current and driftvelocity of electrons in a conductor. Usethis relation to explain how the resistanceof a conductor changes with the rise intemperature? All Indio 2013

o Chopterwise eBSE Solved Papers PHYSICS

13. Plot a graph showing the variation ofresistance of a conducting wire as afunction of its radius. Keeping the lengthof the wire and its temperature asconstant. Foreign 2013

14. Two materials Si and Cu, are cooledfrom300 K to 60 K. What will be the effect ontheir resistivity? Foreign 2013

15. When electrons drift in a metal fromlower to higher potential, does it meanthat all the free electrons of the metal aremoving in the same direction? Delhi 2012

16. Show on a graph, the variation ofresistivity with temperature for a typicalsemiconductor. Deihl 2012

17. Two wires of equal length, one of copperand the other of manganin have the sameresistance. Which wire is thicker?All India 2012

18. Define resistivity of a conductor. Write itsSI unit. All India 2011

19. A wire of resistance 8 Q is 0bent in the form of a circle. A BWhat is the effectiveresistance between theends of a diameter AB? Delhi 2010

20. Twoconducting wires X and Y of samediameter but different materials arejoined in series across a battery. If thenumber density of electrons in X is twicethan that in Y, then find the ratio of driftvelocity of electrons in the two wires.All India 2010

21. Two identical slabs, of a given metal, arejoined together, in two different ways, asshown in figures (a) and (b).

~~@(a) (b)

What is the ratio of the resistances ofthese two combinations? Deihl 2010C


22. The three coloured bands, on a carbonresistor are red, green and yellow,respectively. Write the value of itsresistance. All Indio 2009C

o 2 Marks Questions23. A metal rod of square cross-sectional area

A having length I has current I flowingthrough it when a potential difference ofV volt is applied across its ends (figure I).Now the rod is cut parallel to its lengthinto two identical pieces and joined asshown in figure II. What potentialdifference must be maintained across thelength of 2 I so that the current in the rodis still I?I+-- V----+l

~ I ~ ~L....--;---,--,..--r-r-(I) (II)

Foreign 2016

24. Using the concept of drift velocity ofcharge carriers in a conductor, deduce therelationship between current density andresistivity of the conductor. Delhl2D15C

25. Estimate the average drift speed ofconduction electrons in a copper wire ofcross-sectional area 1.0 x 10-7 m 2 carryinga current of 1.5 A. Assume the density ofconduction electrons to be 9 x 1028m -3.

All Indio 2014

Or Estimate the average drift speed ofconduction electrons in a copper wire ofcross-sectional area 2.5 x 10-7 m 2 carryinga current of 1.8 A. Assume the density ofconduction electrons to be 9 x 1028 m -3.

All Indio 2014

Or Estimate the average drift speed ofconduction electrons in a copper wire ofcross-sectional area 2.5 x 10-7 m2

carrying a current of 2.7 A. Assume the

79

density of conduction electrons to be9 x 1028m-3.

All Indio 2014

26. Draw a plot showing the variation ofresistivity of a (i) conductor and(ii) semiconductor, with the increase intemperature. How does one explain thisbehaviour in terms of number density ofcharge carriers and the relaxation time?Delhl2014C

27. Derive an expression for the currentdensity of a conductor in terms of the driftspeed of electrons. Foreign 2014

28. Define mobility of a charge carrier. Writethe relation expressing mobility in termsof relaxation time. Give its SI unit.All Indio 2013C

29. A conductor of length I is connected to aDC source of potential V. If the length ofthe conductor is tripled by graduallystretching it, keeping V constant, how will

(i) drift speed of electrons and(ii) resistance of the conductor be

affected? Justify your answer.HOTS: Foreign 2012

30. (i) You are required to select a carbonresistor of resistance 47kn ± 10%from a large collection. What shouldbe the sequence of colour bands usedto code it?

(ii) Write the characteristics of manganinwhich make it suitable for makingstandard resistance. Deihl 2011: Foreign 2011

31. Plot a graph showing temperaturedependence of resistivity for a typicalsemiconductor. How is this behaviourexplained? Deihl 2011

32. The sequence of coloured bands in twocarbon resistors ~ and R2 is

(i) brown, green, blue and(ii) orange, black, green.

Find the ratio of their resistances.Delhl2010C

80

33. A wire of 15 n resistance is graduallystretched to double its original length. Itis then cut into two equal parts. Theseparts are then connected in parallelacross a 3.0 V battery. Find the currentdrawn from the battery. All India 2009

34. Derive an expression for drift velocity offree electrons in a conductor in terms ofrelaxation time. Deihl 2009

35. (i) Derive an expression for drift velocityof free electrons.

(ii) How does drift velocity of electrons in ametallic conductor vary with increasein temperature? Explain. All India 2009

36. (i) Define the term of drift velocity.(ii) On the basis of electron drift, derive

an expression for resistivity of aconductor in terms of number densityof free electrons and relaxation time.On what factors does resistivity of aconductor depend?

(iii) Why alloys like constantan andmanganin are used for makingstandard resistors? Delhi 2016

o 3 Marks Questions37. Find the relation between drift velocity

and relaxation time of charge carriers in aconductor. A conductor of length Lisconnected to a DC source of emf E. If thelength of the conductor is tripled bystretching it, keeping E constant, explainhow its drift velocity would be affected.Delhi 2015

38. In the two electric circuits shown in thefigure, determine the readings of idealammeter (A) and the ideal voltmeter (V).

- + + -V V

:6V---:I II I

j------ ..

: 6VI

II

~------ ..II!...------..I

(a) (b)

All India 2015

o ehapterwise eBSE Solved Papers PHYSICS

39. (i) Deduce the relation between currentI flowing through a conductor anddrift velocity v d of the electrons.

(ii) Figure shows a plot of current Iflowing through the cross-section of awire versus the time t. Use the plot tofind the charge flowing in 10 secthrough the wire.

J(A)

10

5

O¥---~------~--~10 t (s) All India 2015C

40. Define relaxation time of the freeelectrons drifting in a conductor. How it isrelated to the drift velocity of freeelectrons? Use this relation to deduce theexpression for the electrical resistivity ofthe material. All India 2012

41. (i) Derive the relation between currentdensity j and potential difference Vacross a current carrying conductorof length I, area of cross-section Aand the number density n of freeelectrons.

(ii) Estimate the average drift speed ofconduction electrons in a copper wireof cross-sectional area 1.0 x 10-7 m 2

carrying a current of 1.5 A. [Assumethat the number density ofconduction electrons is 9 x 1028 m-3].

Delhi 2012C

42. A network of resistors is connected to a16 V battery of internal resistance of 1 nas shown in the figure.

40 120

16V 10

(i) Compute the equivalent resistance ofthe network.

(ii) Obtain the voltage drops VAB andVCD' Fareign 2010


43. Calculate the steady current through the2n resistor in the circuit shown in thefigure. HOTS;Foreign 2010

A B

3Q

4Q

6 V 2.8 Q

44. Three resistors Rt, ~ and ~ areconnected in parallel, across a source ofemf E and negligible internal resistance.Obtain a formula for the equivalentexpressions for the current through eachof the three resistors. All Indio 2009C

o 5 Marks Question45. (i) Derive an expression for drift velocity

of electrons in a conductor. Hence,deduce Ohm's law.

(ii) A wire whose cross-sectional area isincreasing linearly from its one endto the others, is connected across abattery of V volts. Which of thefollowing quantities remain constantin the wire?(a) Drift speed(b) Current density(c) Electric current(d) Electric fieldJustify your answer. Deihl 2017

IZI Explanations1. Consider the figure,

v-Since, slope of 1> slope of 2... RJ < R2 (112)

81

~Also, we know that resistance is directlyproportional to the temperature. Therefore,T2> TJ• (112)

2. (i) DE is the region of negative resistance becausethe slope of curve in this part is negative. (112)

(ii) Be is the region where Ohm's law is obeyedbecause in this part, the current varies linearlywith the voltage. (112)

3. The resistivity of a metallic conductor is given byP = Po [l +a(T -To)]

Where, Po = Resistivity at reference temperatureTo= Reference temperaturea= Coefficient of resistivity

From the above relation, we can say that the graphbetween resistivity of a conductor withtemperature is straight line. But, at temperaturesmuch lower than 273 K ( i.e. O°C), the graphdeviates considerably from a straight line as shownin the figure.

fE'at 0.4

a..?:- 0.2:~Ci5

& 0 50 100 150Temperature T (K) -+

4. Conductors contain free electrons. In the absenceof any external electric field, the free electrons arein random motion just like the molecules of gasin a container and the net current through wire iszero. If the ends of the wire are connected to abattery, an electric field (E) will setup at everypoint within the wire. Due to electric effect of thebattery,the electrons will experience a force in thedirection opposite to E.

5. Variation of current versus voltage for the materialGaAs.

NegativeNon-line i resistanceregion I region

I

III

Voltage M(1)

82

6. Graph of resistivity of copper as a function oftemperature is given below (resistivity of metalsincreases with increase in temperature).

EC:'t 0.4~0:.~ 0.2~·wQlrr L-__-+ +-__~-- __

o 50 100 150Temperature T (K) -

7. The term drift velocity of charge carriers in aconductor is defined as the average velocityacquired by the free electrons along the length ofa metallic conductor under a potential differenceapplied across the conductor. (112)

Its relationship is expressed asI

Vd=-neA (1/2)

where, I is current flowing through the conductor,n is concentration of free electrons, e is electroniccharge and A is cross-sectional area.

8. The electrical conductivity (0) of a metallic wire isdefined as the ratio of the current density to theelectric field it creates. Its SI unit is mho permetre (0-1 /m). (1)

9. The resistivity of a semiconductor decreasesexponentially with temperature.

The variation of resistivity with temperature forsemiconductor (Si) is shown in figure below.

l~~a.-g·~S§~"''''Ql-rro

T-Temperature

10. The mobility of charge carriers in a conductor isdefined as the magnitude of drift velocity (in acurrent carrying conductor) per unit electric field.

(1/2)Drift velocity (vdl q't

Jl = =-Electric field (El m

where, or is the average relaxation time and m isthe mass of the charged particle.Its SI unit is m2/V-s or rns " N-1 C

o Chapterwise CBSE Solved Papers PHYSICS

(1)

11. Increasing temperature causes greater electronscattering due to increased thermal vibrations ofatoms and hence, resistivity P (reciprocal ofconductivity) of metals increases linearly withtemperature.

12. Relation between current and drift velocity ofelectrons in a conductor is given by

I=Anevd

where, I = current,A = area of conductor,n = number density of electrons and

vd = drift velocity.with the increase in temperature of a metallicconductor, resistance increases and hence, driftvelocity decreases. (1)

13. Resistance of a conductor of length Iand radius ris given by

IR=p-

1tr2

r-

14. In silicon, the resistivity increases with decreasein temperature.In copper, the resistivity decreases with decreasein temperature.

15. No, the drift speed of electrons is superposed overthe random velocities of the electrons. (1)

16. Refer to Ans. 9. (1)

17. Given that resistance of both the wires are ofequal length, so

RMn = Rcu

PMnlMn _ lcu----- -Pcu-AMn Acu

... (i)

(1) According to the question, both the wires are ofequal length, so IMn = lcu

:. From Eq. (i), we getPMn = PCuAMn Acu

£. = constantA

or

or Acu = Pcu or P cc AAMn PMn

(112)

CHAPTER 3 Current Electricity

18.

We know that copper is better conductor thanmanganin, therefore, copper will have lessresistivity.

i.e. PCu < PMn

So, AMn > Acu (.: P cc A)

That means wire of manganin will be thickerthan that of copper. (1)

The resistivity of the material of conductor isequal to the resistance offered by the conductorof same material of unit length and unitcross-sectional area. The resistivity of a materialof the conductor does not depend on thegeometry of the conductor. SI unit of resistivity isohm-metre (Q -m). (1/2 + 112 = 1)

The resistance of the whole wire is BQ, which isbent in the form of a circle. We have to find theeffective resistance between the ends of diameterAB. Diameter of the circle divides the circle intotwo equal parts. The resistance of each such partwill be ~ = 4Q.

2

(Resistance R cc length of wire /, if length ishalved then resistance will also become half).From the figure, it is clear that both the parts arein parallel combination. So, effective resistancebetween A and B is given by

1 1 1--=-+-RAB R[ R2

1 I 1~ --=-+-RAE 4 4

19.

A----c;::J-B4Q

4~ RAE =--=2Q

1 + 1

20. Given that number density in X

= 2 x Number density in Y

~ nx = 2ny

As current is common for the enrin- CIrcuit

1= nxAx e (vd)x

= nyAy e (vd}Y

Also, the diameters of the wires are sameAx =Ay

(Vdlx = ny

(vd}y n,

i.e.

~

=2=~2ny 2

83

21. In these types of questions, first of all, identify thecombination in which the metal slabs areconnected and then apply the formula forequivalent resistance accordingly.

Let the resistance each of conductor is R.Case I According to Fig. (a) the resistances areconnected in series combination, so equivalentresistance of slab R[ = R + R = 2RCase IT According to Fig. (b), the resistances areconnected in parallel combination, so equivalentresistance

1 1 1 1 2 R- = - + - ~ - '" - ~ R2 = -R2 R R R2 R 2

Ratio of the equivalent resistance in twocombinations is.. .&=~=4 ~ '&=4

R2 (RI2) R2 (1)

22. According to the colour code of resistances.Code for red = 2, Code for green = 5andCode for yellow = 4:. Resistance of the wire = 25 x 104 Q ± 20% (1)

23. From Ohm's law, we have V = IR

~ V=Ipi [.:R= piJ1 ... (i)A A rn

When the rod is cut parallel, and rejoined bylength, the length of the conductor becomes 2 [,whereas the area decrease to ~. If the current

2remains the same, the potential changes as

21 [V = Ip-=4xlp- =4V [Using Eq. (i}l

AI2 A

The new potential applied across the metal rodwill be four times the original potential (V). (1)

As, we know that I = neAvd

Also current density J is given by J=iA (1)

24.

(1)

[-: vd=e;:]

[-:p= n~~](1)

or, J = ~EP

(1)

25. Given, cross-sectional area, A = 1.0 x 10-7 m2

Current, I = 1.5AElectron density, n = 9 x 1028 m"

Drift velocity, Vd =?We know that, I = neAvd (1)

I 1.5~ vd = - = ---=~-----;-:;;------:;-

neA 9 x 1028 x l . 6 x 10 [9 X 1.0 x 10 7

= 1.042 x 10-3 rn/s (1)

OR (Am. = 5xlO-4m/s) OR (Ans = 7.5XlO-4m/s)

84 o Chopterwise eSSE Solved Papers PHYSICS

26. (i) Conductor0.4

0..?;- 0.2's~'(ii<1la:

o 50 100 150

Temperature T (K) -

(ii) Semiconductor

Temperature (T)

, The relation between resistivity and relaxation. m

tune, P=-2-ne=:

In conductors, average relaxation time decreaseswith increase in temperature, resulting in anincrease in resistivity.In semiconductors, the increase in numberdensity (with increase in temperature) is morethan the decrease in relaxation time, the netresult is therefore a decrease in resistivity. (1)

Let potential difference V is applied across aconductor of length I and hence, an electric field Eproduced inside the conductor,

.. E =~ ... (i)I

27.

I+------I--------+l~-e-e-e

Conventionalcurrent

Electriccurrent

II

L--------l+ 111-_----------'Battery

Let n = number density of free electrons

(1)

A = cross-sectional area of conductore = electrons charge

:. Number of free electrons present in length I ofconductor = nAl

:. Total charge contained in length I which cancontribute in current

q = (nAI)e ... (i) (112)

The time taken by free electron to cross thelength I of conductor is

t=~Vd

... (ii)(112)

where, vd = drift speed of electron:. Current through the conductor

I=it

1= (nAI)et

(nAI)e= (~) = neAvd

C densi () I ne AVd:. urrent ensity J =- =--= nevdA A

28.

.. J = nevd i.e. F oc Vd (1)

Thus, current density of conductor is proportionalto drift speed.Mobility of a charge carrier is defined as the driftvelocity of the charge carrier per unit electric field.It is generally denote by 11.

11 = vd

EThe 51 unit of mobility is m2 V-I S-l . (1)

Drift velocity in term of relaxation time:-eE

Vd=--'tm

In magnitude,

or

eEvd =-'t

mvd = e'tE m

e't11 =-

m (1)

29. When a wire is stretched, then there is no changein the matter of the wire hence, its volume remainsconstant.

The potential V = constant, l' = 31


. V-(i) Drift speed of electrons =--

ne/pwhere, n is number of electrons, e is charge onelectron, I is the length of the conductor and p isthe resistivity of conductor.

1v DC - (.: Other factors are constants)I

So, when length is tripled, drift velocity getsone-third. (1)

Resistance of the conductor is given asI

R=p-A

Here, wire is stretched to triple its length, thatmeans the mass of the wire remains same inboth the conditions.Mass before stretching = Mass after stretching(Volume x Density) before stretching

= (Volume x Density) after stretching(Area of cross-section x Length) beforestretching = (Area of cross-section x Length)after stretching

(.: Density is same in both cases)Alii = A2/2All = A2 (31)

(": Length is tripled after stretching)AlA2=-3

i.e. when length is tripled area of cross-sectionis reduced to ~.

3I'

R=p-A'31

=Pjf3

I=9p-=9R

AThus, new resistance willoriginal value.Given, resistance = 47 kn ± 10%

= 47 x 103 n ± 10%

(ii)

Hence,

(1)be 9 times of its

(1)

30. (i)

:. Ist colour band should be yellow as code forit is 4.IInd colour band should be violet as code for itis 7.IIIrd colour band should be orange as code forit is 3.Nth colour band should be silver becauseapproximation is ± 10%

85

Violet 1Df OrangeYellow _ _ Silver-E) )-(I)

(ii) Two properties of manganin are(a) low temperature coefficient ofresistance.(b) high value of resistivity of material of

manganin make it suitable for making astandard resistor. (I)

31. To plot the graph between the two quantities. firstof all identify the relation between them.

Since, resistivity of material of conductor (P) is

given by, p = ~ne2t

where, n = number density of electrons,and t = relaxation time.With the rise of temperature of semiconductor,number density of free electrons increases, whereast remains constant and hence resistivity decreases.

(I

32.

T-Resistivity of 0 semiconductor decreoses

ropidly with temperoture

According to colour codes, resistance of two wiresare(i) Code of brown = 1

Code of green = 5Code of blue = 6

RI =15x106n ± 20%

(ii) Code of orange = 3Code of black = 0Code of green = 5

R2 = 30 x 105n ± 20%

. . R 15x1~:. Ratio of resistances. -.!. = 5 = 5R2 30x10

RI = 5R2

86 o ehapterwise eSSE Solved Papers PHYSICS

33. Let original cross-sectional area and length of15 n resistance are A and I after stretching theybecome A' and 1', respectively.

Initial resistance, R = pi. ~ 15 = pi. ...(i)A A

:. In case of stretching, volume of the wireremains same, so

AI= A'I'

but 1'=21 ~ A'=~ ... (ii)2

:. Resistance after stretching

, I' (21) (A)R = PA' = P AI2 = 4 Pi

R' = 4 x 15 [from Eq. (i))Now, resistance, R' = 60 n (1/2)

After dividing into two parts, resistance of eachpart = 30 n., Effective resistance after connecting them intoparallel combination

30R<ff="2=15n

(1/2)

': Applied potential difference, V = 3 VV., Current drawn from the battery, 1 = -R

(from Ohm's law)

1=2 ~ 1=~A15 5

34. When a conductor is subjected to an electric fieldE, each electron experiences a forceF = - e E, and free electron acquires anacceleration

F eEa=-=-- ... (i)m m

where, m = mass of electron, e = electronic chargeand E = electric field.Free electron starts accelerating and gains velocityand collide with atoms and molecules of theconductor. The average time difference betweentwo consecutive collisions is known as relaxationtime of electron and

t='tI+'t2+ ..·+'tn ... (ii)n (1)

where, 'tI, 't2, ..., 'tn are the average time differencebetween 1st, 2nd, ... , nth collisions....VI ' V2' , Vnr are velocities gained by electron inIst, 2nd , nth collisions with initial thermalvelocities uI' u2, .•• , un' respectively... VI =UI +a'ti

(1)

The drift speed vd may be defined as__V..!..I_+_V~2_+_·_··_·+_V.!!,nVd -

n_ (UI + U2+ ....+ un) + a ('ti + 't2 + ...+ 'tn)

Vd - ~'---"-----"-'---'-..!..---'----''-'.n

_(\11 +U2+····+Un) a('ti +'t2+···+'tn)Vd - + ---,,-,-_~ __ ---,,c.:.n n

Vd = 0 + a t ['.'Average thermal velocityin n collisions = 0)

Vd = - (e:) 't [from Eq. (w)This is the required expression of drift speed offree electrons.

35. (i) If t be the average time between twosuccessive collision, the distance during thisperiod is

I =~a('t)2 =~(eE) ('t)222m (ll

Drift velocity, I 1 eE't2vd =-=---'t 2m 't

eEvd =-'t2m (1)

(ii) As the temperature of a conductor isincreased, the thermal agitation increases andthe collisions become more frequent. Theaverage time 't between the successivecollisions decreases and hence the drift speeddecreases. Thus, the conductivity decreasesand the resistivity of the conductor increases.

(1)

36. (i) Drift Velocity Refer to Ans. 7.The drift velocity of electron is of the order oflO-4m/s.

(ii) Specific resistance or resistivity of the materialof a conductor is defined as the resistance of aunit length with unit area of cross-section ofthe material of the conductor.The unit of resistivity is ohm-metre or nom.

Since, we know that R = pi.A

RA~. p=- .... (i)I

From Ohm's law, V=IR~El=neAvdREl eE't~ R=--andvd=-

neAvd m


Elxm mlSO,R 2

ne2 AEt ne At

Substituting the value of R =~ in Eq. (i),ne At

We have, p = (~).~ne At 1

:.Resistivity of the material, p=+ne t

From the above formula, it is clear thatresistivity of a conductor depends upon thefollowing factors:

(a) pee~, i.e. the resistivity of material isn:

inversely proportional to the numberdensity of free electrons (number of freeelectrons per unit volume). As the freeelectron density depends upon the natureof material. so resistivity of a conductordepends on the nature of the material.

(b) peel I r. i.e. the resistivity of a material isinversely proportional to the averagerelaxation time t of free electrons in theconductor. As the value of t depends on thetemperature as temperature increases, tdecreases, hence p increases. (1)

(ill) Alloys like constantan and manganin are usedfor making standard resistor because theresistivity of these alloys here weak dependenton the temperature. (1)

37. Refer to Ans. 36(ii). (1)

The conductor connected to DC source of emf E isshown in the figure below:

CL._c.;,.,o_n_du_c_Io_r -1liE

Suppose, initial length of the conductor. l; = 10.New length. l, = 310 (1)

we know that,Drift velocity. vd cc E 0 (electric field)Thus. (Vd), = (E 0), = Ell,

(vd)i (E 0) i E Ili

=~=JL=~l, 310 3

(vd) iThus, (Vd), =--

3Thus, drift velocity decreases three times. (1)

87

38. In first circuit,Reading of ideal voltmeter = 6YNet potential difference = 9 + 6 = 15 YTotal resistance = 1+ 1=20

Current in ammeter =.!:: =~= 7.5AR 2 (1)

In second circuit,Reading of ideal voltmeter = 6YNet potential difference = 9 - 6 =3Y (1)

Total resistance = 1+ 1= 20

Current in ammeter =.!:: =!=1.5AR 2 (1)

39. When an electric field is applied across aconductor, then the charge carriers inside theconductor move with an average velocity which isindependent of time. This velocity is known asdrift velocity (vd). (1)

(i) Relationship between current (1) and velocity(vd) I=neAvd

where, ne = amount of charge inside theconductor and A = area of cross-section ofconductorTotal number of free electrons in a conductorPQ of length I, cross-sectional area A having nfree electrons per unit volume is

N = n x volume of conductor PQor N = nAl

Time I in which an electron moves from P to Q,all N free electrons pass through cross-sectionQ.

I=~Vd

where Vd is the drift velocity of electrons in theconductor. So electric current flowing throughconductor is given by

q Ne nAleI=-=-=--~I= neAvd

I I I I Vd

This gives the relation between electric currentand drift velocity. (1)

(ii) Area under I-I curve on r-axis is charge flowingthrough the conductor

Q= ~ x 5x 5+ (10+ 5)x 5=87.5C2 m

40. Relaxation Time The average time differencebetween two successive collisions of driftingelectrons inside the conductor under theinfluence of electric field applied across theconductor, is known as relaxation time.Drift speed and relaxation time

eE'tVd =---

m (1I2lwhere, E = electric field due to applied potentialdifference

't = relaxation timem = mass of electrone = electronic charge

Electron current, I = - ne A vd

I = - neA ( _ e!'t)

)J II 1= ne2A't (~)~..• 'II ~I \ J m I

V mI I m~ -=--=p-=R ~p=--I ne2A't A ne2't

88

[II ,

This is the required expression.

41. (i) The current in the conductor having length 4cross-sectional area A and number density n is

... (i)1= neAvd

Electric field inside the wire is given by

E=~I

If relaxation time is r. the drift speede'tE

vd=-m

where, m = mass of electron't = relaxation chargee = electronic charge

E = electric field.Put the value of Eq. (i), we get

2I=ne'tAE

m~

(ii)

From Eqs. (ii) and (ill), we get1= ne2'tAV ~ J =!.. = ne2'tV

ml A mlGiven, 1= 1.5 A, n = 9 x 1028 m-3,

A = 1 .0 X 10-7 m2

IVd=-

neA1.5

:. Vd = ----,;;;------",;-------;;-9 X 1028 x 1.6 x 10 19 X 1.0 xlO 7

~Vd = 1.04 x 10-3 rn/s

(l/2l

(l/2l

[-: E = T](1I2l

... (ii)

... (ill)


42. To calculate the equivalent resistance of complexnetwork (network having multiple branches).calculate the equivalent resistance of smaller part ofnetwork and finally calculate the equivalentresistance of the network.m

(i) ':4 il and 4 il are in parallel combination.:. Equivalent resistance,

I IRAB =- +-4 4

I 2-- = - ~ RAB = 2ilRAB 4

Similarly, equivalent resistance of 12 il and6 il is

I I I I 1+2- = - + - ~- = -- ~ RBC = 4 ilRBC 12 6 RBC 12

Now, the circuit can be redrawn as shown infigure below

m

Now, 2il, I il and 4il, I il are in seriescombination.:. Equivalent resistance of the network

R.,q = zn + Iil+ 4il+ Iil= an tn(ii):. Current drawn from the battery,

I = ~ = 16 = 2AR 8

This current will flow from A to Band C to D.So, the potential difference in between AB andCD can be calculated asNow, VAB = !RAB = 2 x 2 = 4 V (lland VCD = I RCD = 2 x 2 = 8 V rn

43. To calculate the current through a particularresistance, first we have to find the potentialdifference across that resistance.

m In DC circuit, capacitor offers infinite resistance.Therefore, no current flows through capacitor andthrough 4 il resistance, so resistance will produceno effect.:. Effective resistance between A and B

2x 3RAB = -- = 1.2 il2+ 3

[I I I RR].: R = RI + R2 ~ R = RI ~ ~2 rnm


Total resistance of the circuit = 1.2+2.8 = 4(2(.: These two are in series)

Net current drawn from the cell,

1= V = ~ = ~ = 1.5AR (total resistance) 4 2 (112)

:. Potential difference between A and B

VAB = IRAB = 1.5 x 1.2 ~ VAB = 1.80 V (112)

Current through 2(2 resistance,

I' = VAB = 1.8 = 0.9 A2(2 2

44. Let the equivalent resistance of parallelcombination of RJ, R2 and R3 is R.

1 1 1 1-=-+-+ -R RJ R2 R3

.!. = R2R3+ RJR3 + RJR2

R RJR2R3

R = RJR2R3RJR2 + R~3 + R3RJ

Effective current, I = ~R

1= E(RJR2 + R2R3 + R3RdRJR2R3

45. (i) Let vd be the drift velocity.

(1y.)

(1Y.)

89

(1)

Electric field produced inside the wire is,

E =.!:: ... (i)I

Force on an electron = - E e (1)

Acceleration of each electron = _ Eem

[.: from Newton's law, a = F / m]where, m is mass of electron.

Velocity created due to this acceleration =Ee t ,m

where, 't is the time span between twoconsecutive collision. This ultimately becomesthe drift velocity in steady state.

Ee e VSo vd =-'t =- 't X -. [From Eq. (i)]m m I

We know that current in the conductori = n e A vd (n is number of free electrons in aconductor per unit volume)

. e V . n e2A 't Vl=neAx-'t- ~ 1=----

m I m I

. VZ=-

R [.: R _ m I ]

- n e2A't

i oc V

This is Ohm's law. (1)

(ii) The setup is shown in the figure. Here, electriccurrent remains constant throughout thelength of the wire. Electric field also remainsconstant which is equal to .!::.

I

(1)

Current density and hence, drift speed changes.

[TOPIC 2] Kirchhoff's Laws, Cell and TheirCombinations

2.1 CellA device Rwhich is used C 0to maintain asteady currentin an electriccircuit is ~lealfe'd' cell or1r l~ith3l-..Dt,' .e9 l>89nY,rl,._1n;lJt5J; "~ Electrolyteceu, I has Celltwo ~~~ctrodesRositive (f) and negative (N) as shown infiB '2Jl~~()ve, ');1 •r· .

f n?' t' I ']:)', II '", '

Terms Related to CellEMF of a CellIt is' the maximum potential differencebetween two terminals of circuit, whencircuit is open.EMF of the cell, E = W,

q

Internal Resistance (r).. - I - .

Internal resistance of a cell is defined as theresistance offered by the electrolyte of thecell to the flow of current through it. It isdenoted by t. Its SI unit is ohm (Q),

Terminal Potential Difference (V)The maximum potential difference betweentwo terminals of circuit when 'the circuit isclosed it, is known as terminal voltage orterminal potential difference (V) of cell,

Relation between" R, E and V

r = R(~-1) '" (i)

where, t = internal resistance,R = external resistance,E = emf of cell,V = terminal voltage of cell

Also, V = E - Ir ".(ii)

and V=(R!r)R or V= (1:~) ...(iii]The terminal voltage increases with theincrease of external resistance R,

Charging of a CellDuring charging of the cell, V = E + Ir

I [-~~~~

So, V< E, when current is drawn from the cell,i.e. discharging,and V> E when charging of cell takes place,

2.2 Cells in Series and ParallelSeries CombinationIn series combination, current is given by,

1= nE(R+nr)

!-E~:~~-:4-LE;----r;----1I .--vYVVVTTl, I I

~------------..! ~------------..!

IR I

where,

Parallel CombinationIn parallelcombination,current is givenby;

1= mE(r+mR)


,-£---------:: I 1 r1 I

: I :~----- I

,-----------,I E I: 12 r2 I

: II

--

1

R I

Mixed GroupingIt consists of m-rows in parallel combinationsuch that each row contains n-cells of each ofemf E and internal resistance r. then current inthe circuit is given by

1= mnEmR+nr

--- n-cells ---E1r E1r E1rr-j 1- - - - - --I

E1r E1r E1rr-jl-------I

E1r E1r E1rr-jl-------I

~~ ------------------1

R

and maximum current is drawn from thebattery, when external resistance matches withnet internal resistance i.e.

nr nE mnE mER=- and 1 =--=--=-

m max 2(:) 2nr 2r

2.3 Kirchhoff's LawsKirchhoff has given two rules based onconservation of electric charge and of energy,these are known as Kirchhoff's laws.

Kirchhoff's First Law (Junction Rule)The algebraic sum of electric currents at anyjunction of electric circuit is equal to zero i.e. thesum of current entering into a junction is equalto the sum of current leaving the junction

91

11

D=O,14=11+12+13Junction law supports law of conservation of chargebecause this is a point in a circuit which cannot actas a source or sink of charge(s).

. ).L:JNOTE The current flowing towards the junction. Fief

conductors is considered as positive anq. t~!'! h-I,I{Ientflowing away from the junction is tak n a ne~a i'jk

~ 1 9-,

Kirchhoff's Second Law (~o~~ R9!~~2o~In any closed loop of electrical circuit, the ,etilIC braic

Go'9 rur,sum of emf's of cell and the product of currents andresistance is always equal to zero i.e.

UV=Oor~E=~IR

Kirchhoff's second law supports the law ofconservation of energy, because the net change inthe energy of a charge, after completing a closedpath must be zero.NOTE Sign convention for Kirchhoffs second law.

V= +E V=-E

. 1 R~

V= -1R

1 R~

V= +1R

An Application based onKirchhoff's LawLet us consider a circuit diagram

A I B11

R2

C

1212

F

11 I E1 R1

D~~~~NV~-~11 + 12 R3

f2E

92

Current distribution is shown in givencircuit using Kirchhoff's first law.Now, apply loop rule for the mesh DCBAD

E] -(I] + I2)R3 -I] (R, + R2) = 0

~ -IdR, + R2 + R3) + I2(-R3) + E, = 0 ... (i)Similarly, applying loop rule for the meshCDEFC

E2-I~4-(I, +I2)R3-I~5=O

~ IdR3) + I2(R3 + R4 + R5) = E2

~ I,(R3)+I2(R3+R4+R5)-E2=O ... (ii)By Eqs. (i) and (ii), we can calculate I] and 12,

PR·EVIOUS YEARS'EXAMINATION QUESTIONSTOPIC)2o 1 Mark Questions

1. The plot of the variation of potentialdifference across a combination of threeidentical cells in series versus current isshown below. What is the emf andinternal resistance of each cell?

6V

1V

1-+ 1 A

2. Two identical cells, each of emf E, havingnegligible internal resistance, areconnected in parallel with each otheracross an external resistance R. What isthe current through this resistance?All India 2013

3. A 10 V battery of negligible internalresistance is connected across a 200 Vbattery and a resistance of 38 n as shownin the figure. Find the value of thecurrent in circuit. Deihl 2013


10V

r'- J~~~10V

4. The emf of a cell is always greater thanits terminal voltage. Why? Give reason.Oelhi 2013

5. A cell of emf E and internal resistance rdraws a current I. Write the relationbetween terminal voltage V in terms of E,I and r. Delhi 2013

6. Three cells of emf E, 2E and 5E havinginternal resistances r, 2r and 3rrespectively are connected across avariable resistance R as shown in thefigure. Find the expression for thecurrent. Plot a graph for variation ofcurrent with R.

E 2E 5EC~HR All India 2010C

o 2 Marks Questions7. A battery of emf 12 V and internal

resistance 2 n is connected to a 4 nresistor as shown in the figure.

'----"I>M----{ A40

(i) Show that a voltmeter when placedacross the cell and across theresistor, in turn, gives the samereading.

(ii) To record the voltage and the currentin the circuit, why is voltmeterplaced in parallel and ammeter inseries in the circuit? All India 2016


8. Use Kirchhoffs rules to determine thepotential difference between the pointsA and D. When no current flows in thearm BE of the electric network shown inthe figure below:

30F.-~~--~Er-------,D

R

CDelhi 2015

9. Use Kirchhoffs rules to determine thevalue of the current IIflowing in thecircuit shown in the figure.

300 11

11

.1'I

80V 20 0 12 Delhi 2013C

10. In the electric network shown in thefigure, use Kirchhoffs rules to calculatethe power consumed by the resistanceR=4Q.

A

R=40B~--~----~~r---~D

11. Two cells of emfs 1.5 V and 2.0 V havinginternal resistances 0.2 Q and 0.3 Qrespectively are connected in parallel.Calculate the emf and internalresistance of the equivalent cell. Delhi 2014

12. A cell of emf E and internal resistance ris connected across a variable resistor R.Plot a graph showing variation ofterminal voltage V of the cell versus thecurrent 1. Using the plot, show the emfof the cell and its internal resistancecan be determined. All India 2014

93

13. Distinguish between emf (e) and terminalvoltage (V) of a cell having internalresistance r. Draw a plot showing thevariation of terminal voltage (V) versus thecurrent (1) drawn from the cell. Using thisplot, show how does one can determine theinternal resistance of the cell? All India 2014 C

14. A battery of emf E and internal resistancer when connected across an externalresistance of 12 Q, produces a current of0.5 A. When connected across a resistanceof 25 Q, it produces a current of 0.25 A.Determine (i) the emf and (ii) the internalresistance of the cell. All India 2013C

15. A cell of emf E and internal resistance r isconnected to two external resistap,ceaand Rz and a perfect ammeter. Tlie .currenti~ the.circuit is measured in four different'3situations:

(i) Without any external resistanc 'Itt~, Tcircuit

(ii) With resistance s;only(iii) With ~ and Hz in series combination(iv) With ~ and Hz in parallel combinationThe currents measured in the four cases are0.42 A, 1.05 A, and 4.2 A, but notnecessarily in that order. Identify thecurrents corresponding to the four casesmentioned above. Delhi 2D12

16. A battery of emf 10 V and internalresistance 3 Q is connected to a resistor.If the current in the circuit is 0.5 A, find

(i) the resistance of the resistor(ii) the terminal voltage of the battery

Delhi 2012C

17. The network PQRS, 2000

shown in the circuit ~Rdiagram, has thebatteries of 4 V and 5 V mA 9' 6005 V and negligible '1"t::>

internal resistance. P QA milliammeter of 20 Q 4 Vresistance is connected between P and R.Calculate the reading in the milliammeter.All India 2012C

94

18. Calculate the current drawn from thebattery in the given network.

Rs = 2Q

+ -4V

All India 2012C

19. In the given circuit, assuming point A tobe at zero potential, use Kirchhoffs rulesto determine the potential at point B.

2V1 A 0 2,A I~ B·

RALI'I j2n JR'1 V C 2A

All India 2011

20. Using Kirchhoffs rules in the givencircuit, determine

(i) the voltage drop across theunknown resistor Rand

(ii) the current 12 in the arm EF.O.SA 3V 2QA B

RAll India 2011

21. A straight line plot showing the terminalpotential difference (V)of a cell as afunction of current (1) drawn from it isshown in the figure. Using this plot,determine

(i) the emf and(ii) internal resistance of the cell.

Delhi 2011C

22. Two cells of emf 2E and E and internalresistances 2r and r respectively, areconnected in parallel. Obtain theexpressions for the equivalent emf andthe internal resistance of thecombination. All India 2010C

IZl Chapterwise CSSE Solved Papers PHYSICS

23. Three cells of emf E, 2E and 5E havinginternal resistances r, 2r and 31', variableresistance R as shown in the figure. Findthe expression for the current. Plot a graphfor variation of current with R.

r+l+--tl t----jE, r SE, 3r 2E, 2r

'-- ----' All India 2010C

24. A cell of emf E and internal resistance r isconnected across a variable resistor R. Plota graph showing the variation of terminalpotential V with resistance R. Deihl 2009

25. Plot a graph showing the variation ofterminal potential difference across a cellof emf E and internal resistance r withcurrent drawn from it. Using this graph,how does one determine the emf of the cell?Oelhl2009C

IZJ 3 Marks Questions26. A cell of emf E and internal resistance r is

connected across a variable load resistor R.Draw the plots of the terminal voltage V.versus (i) resistance Rand (ii) current 1.It is found that when R = 4 n, the currentis 1 A and when R is increased to 9 n, thecurrent reduces to 0.5 A. Find the values ofthe emf E and internal resistance r.All India 2015

27. Using Kirchhoffs rules, determine thevalue of unknown resistance R in thecircuit, so that no current flows through4 n resistance. Also, find the potentialdifference between points A and D.

1 Q EF r--.rvvW'"--.------~D

1Q

C

4QR

~B

9V 3VA

Deihl 2012


28. Calculate the value of the resistance R inthe circuit shown in the figure, so that thecurrent in the circuit is 0.2A. What wouldbe the potential difference between pointsAandB?

6V SO

100f C

1S0 SO2V

00.2A R A All India 2012

29. In the circuit shown, ~ = 4n,R2 = R3 = 15n, R4 = 30 nand E = 10V.Calculate the equivalent resistance of thecircuit and the current in each resistor.

A

E Ra

11 Delhi 2011

30. State Kirchhoffs rules. Use these rules towrite the expressions for the currents11,12 and 13 in the circuit diagram shownin figure below.

I E1 = 2 V (1 = 4 01 I~--.N~~---,

I

12 E2 = 1 V (2 = 3 0II

1a E3 ~ 4 V (3 = 20I

31. State Kirchhoffsrules. ApplyKirchhoffs rulesto the loopsACBPAandACBQA to writethe expressionsfor the currentsII' 12 and 13 inthe network.All India 2010

cR = 120

95

32. State Kirchhoffs rules. Apply these rulesto the loops PRSP and PRQP to write theexpressions for the currents II' 12 and 13 ingiven circuit. All India 2010

2000 12r-----~~~----~R11

SV 600

4V

r------+~V r.----------,33. Write any twofactors on whichinternal resistanceof a cell depends.The reading on ahigh resistancevoltmeter, when acell is connectedacross it, is 2.2 V. When the terminals ofthe cell are also connected to a resistance of5 n as shown in the circuit, the voltmeterreading drops to 1.8V. Find the internalresistance of the cell. All India 2010

E~---+"-i -

R=SO K

34. (i) State Kirchhoffs rules.(ii) A battery of 10 V and negligible

internal resistance is connectedacross the diagonally opposite cornersof a cubical network consisting of 12resistors each of 1 n resistance. UseKirchhoffs rules to determine(a) the equivalent resistance of the

network and(b) the total current in the network.

All India 2010

35. Twocells of emf E1 , E2 and internalresistances r1 and r2 respectively are

mected in parallel as shown in the figure.E1. (1

A-----<I

?--- .•aI

36.

Deduce the expressions for(i) the equivalent emf of the

combination.(ii) the equivalent resistance of the

combination and(iii) the potential difference between the

points A and B. Foreign 2010

Calculate the steady current through the2n resistor in the circuit shown in thefigure below. 2n

A B

3nI-I---~NV

211F 4 n

6V z.e nForeign 2010

37. State Kirchhoffs rules. Use Kirchhoffsrules to show that no current flows in thegiven circuit.

~:8~ Foreign 2009

o 5 Marks Questions38. (i) State the two Kirchhoffs laws.

Explain briefly, how these rules arejustified?

(ii) The current is drawn from a cell ofemf E and internal resistance rconnected to the network of resistorseach of resistance r as shown in thefigure. Obtain the expression for(a) the current draws from the celland (b) the power consumed in thenetwork. Delhi 2017

A B

E,r

o ehopterwise eBSE Solved Papers PHYSICS

o Explanations1. According to question, maximum potential of three

cells (cells in series) each of emf E is given in graph(Lc.6V)So, 3E = 6 V :::} E = ~ = 2 V

3Internal resistance of three cells each of resistancer can be calculated as

V = 1 x 3r [all are in series]3r = ~ = ~ :::} r = 2n

I 1 (1)

2. The cells are arranged as shown in the circuitdiagram given below.

RAs the internal resistance of cells is negligible, sototal resistance of the circuit = R

So, current through the resistance, I = !R (1)

(In parallel combination, potential is same as thesingle cell)

3. Since, the positive terminal of the batteries areconnected together, so the equivalent emf of theba tteries is given by

E = 200 - 10 = 190 V.Hence, the current in the circuit is given by

I=!= 190 = 5AR 38 (1)

4. The emf of a cell is greater than its terminal voltagebecause there is some potential drop across the celldue to its small internal resistance. (1)

5. When a current 1 draws from a cell of emf E andinternal resistance r, then the terminal voltage isV=E-~ m

6. Here E\ =E, E2=-2E and E3=5E, r\ =r, r2=2rand r3= 3rEquivalent emf of the cell is E =E\ + E2 + E 3

=E-2E+5E=4EEquivalent resistance

=r\ + r2+ r3+ R=r + 2r + 3r+ R=6r+ R

:. Current, I =~6r+R

The graph for variation ofcurrent I with resistance R isshown above. (1)

iLo R(ohm)-+


7. According to question,

12V,2n

'-----'VIIVWII'v--{ A4n(i) Net current in the circuit = !3 = 2A

6Voltage across the battery,

Vb =12- z x 2= 8VVoltage across the resistance

Vr = IR = 2 x 4 = 8 V (1)(ii) In order to measure the device's voltage for a

voltmeter, it must be connected in parallel tothat device. This is necessary because device inparallel experiences the same potentialdifference. An ammeter is connected in serieswith the circuit because the purpose of theammeter is to measure the current through thecircuit. Since, the ammeter is a low impedancedevice. Connecting in parallel with the circuitwould cause a short-circuit, damaging theammeter of the circuit. (1)

8. Consider the given figure,

I an E IF.-~~-~-----.D

1VL Rj

2n3V

~ ~6V 4V

Applying Kirchhoffs second law in mesh AFEBA

21-1+31-6=0(Since, no current flows in the arm BE of thecircuit)

R

I

c

51= 7I=!..A

5Kirchhoff's second

... (i)

ApplyingAFDCA

31 + R1 - 4 - 6 + 2I -1 = 051 + R1 =11

law in mesh

... (ii)

Now, substitute the value of I from Eq. (i) toEq. (ii). we get

5X!..+RX!..=115 5

7+ 7R=115

7 R =45

R = 20 n.7 (1)

For potential difference across A and D,along AFD

7 7VA - - x 2+ 1 - 3x - = Vv

5 514 21

VA - - + 1 - - = VD5 5

14 21~ VA - Vv = - + - - 1

5 5(VA - VD) = 7 -1 = 6 V

9. In closed mesh ABeD

Ilrl + (II + 12) R =12211+ 4(11 + 12) = 12211+ 411 + 4[2 = 12

611 + 412 =12311 + 212= 6

In closed mesh BDEF

(II + I2)R = 6

(II + 12)4 = 6211+ 212= 3

On solving Eqs. (i) and (ii), we get r;' = 3A10. According to question,

(1)

... (i)

... (ii

A-(:t"2.0V

E r + E r.Equivalent emf = I 2 2 I

rl + r2

i.e. =(1.5xO.3)+ (2xO.2)0.2+ 0.3

0.45+ 0.40.5

= 0.85 =1.7 V0.5 (1)

98·

Equivalent internal resistance=~= 0.2xO.3

rl+r2 0.2+0.3

= 0.06 = ~ = 012 00.5 50

11. We know that, V = E - IrThe plot between V and I is a straight line ofpositive intercept and negative slope as shown infigure below.

v 1

I

(I) The value of potential difference correspondingto zero current gives emf of the cell. (1)

(ii) Maximum current is drawn when terminalvoltage is zero, so V = E - Ir

~ O=E-Imax,

~ r=~ (1)i.:

12. Difference between emf(f) and terminal voltage (V)

Terminal voltage---------------------S.No. Emf

It is the poten tialdifference betweentwo terminals ofthe cells when nocurrent is flowing

r~c;---IlT:r'7,t'f-hr-o-,u..1I!h.it.~Jl.". --'4r

2. It is the cause.

It is the potentialdifference betweentwo terminals whencurrent passesthrough it.

1.

It is the effect.

-:::.~ Q)

.- OJE ro~gI

CurrentNegative slope gives internal resistance.

13. RI = 120, R2=250

II =0.5A, 12=0.25AFor the 1st case

E E E-6r=--RI =--12 ~ r=--

II 0.5 0.5.. , (i)

o Chopterwise CSSE Solved Papers PHYSICS

(1)

Now, for the 2nd caser=~-25; r=E-6.25

0.25 0.25Compare the Eqs. (i) and (ii), we get

E-6=E-6.25

0.5 0.250.25E-1.5= 0.5E-1125

~ -0.25E = -1.625E=1.625

0.25~ E=6.5VPutting the value of E in Eq (i), we get

6.5-6 0.5r=----=--

0.5 0.5

r=10 (1)

... (ii)

(1)

~14. The current relating to corresponding situations

are as follows:(i) Without any external resistance

EII =-

RIn this case, effective resistance of circuit isminimum so current is maximum. (112)Hence, II =4.2A.

(ii) With resistance Rlonly1,=_E_- r+RI

In this case, effective resistance of circuit ismore than situations (i) and (iv) but less thanSo,I2=1.05A. (1/2)

(iii) With RIand R2in series combinationE

I3=-~--r+RI +R2

In this case, effective resistance of circuit ismaximum so current is minimum.Hence 13 =0.42A. (112)

E(iv) 14r+ RIR2

RI + R2In this case, the effective resistance is morethan (i) but less than (ii) and (iii). So, 14=1.4A.

(112)

(2)

15. GivenE=10V, r=30,I=o.5ATotal resistance of circuit

R+ r=~=.!.Q.= 2001 0.5

(i) External resistance R=20-r=20- 3=170(ii) Terminal voltage V=IR= 0.5xI7=8.5V


16. The given diagram is shown below20011

5V 6011

4VApplying Kirchhoff's second law to the loop PRSP

-I3x20-I2x200+5=0

413 + 4012 = 1 ... (i)

(1)For loop PRQP,

-2013 - 6011 + 4 = 0513+1511 =1

Applying Kirchhoff's first law13 = II + 12 ·... (iii)

From Eqs. (i) and (iii), we have,411 + 4012 = 1

On solving, we get13 = ~A = 1l00°mA

172 172

12 = 4000 mA, II = 39000 mA215 860 (1)

17. The given circuit can be redrawn as given below

... (ii)

'--_"":+!:...j r=-------( }-----'

4V

Here, RI = R4 ~ ~=~R5 R3 2 4

Wheatstone bridge is balanced. So, there will nocurrent in the diagonal resistance R2 or it can bewithdrawn from the circuit. The equivalentresistance would be equivalent to a parallelcombination of two rows which consists of seriescombination of RI and R5 and R4 and R3,respectively.

1 1 1 1 1- =--+ -- =-+-R 1+ 2 2+ 4 3 6

R = 18 = 2119 (1'12)

992

.. I = ~ = ~ = 2A or I = 2AR 2 (112)

18. By Kirchhoff's first law at DIoe =IA (IDC + 1= 2)

1A o 2A 2VIIII B

1 A

R 211 R,

IIIIA C 2A1 V (1)

AiongACDBA, VA + I V + 1 x 2- 2= VBBut VA = 0, VB = 1 + 2 - 2 = 1V

VB =1 V (1)

19. (i) Applying Kirchhoff's second rule in the closedmeshABFEAVB - O.Sx 2 + 3 = VA ~ VB - VA = - 2

V = VA - VB = + 2YPotential drop across R is 1 V as R, EF andupper row are in parallel. (1)

OrPotential across AB = potential across EF

3- zx 0.5 = 4- 212212 = 2A

Potential across R = potential across AB= potential across EF

=i3-2x0.5=2Y(ii) Applying Kirchhoff's first rule at E.

0.5+ 12 = I " . 'Hi'

where, I is current through R.Now, Kirchhoff's second rule in closed meshAEFB, r.E + r.IR = 0 ·.orli 'J j .s

- 4+ 212- O.5x 2+ 3 = 0212-2=0 ~ I2=IA

The current in arm EF = 1A (1)

20. (i) The value of potential differencecorresponding to zero current gives the emf ofcell. This value is 1.4 volt. (1)Maximum current is drawn from the cellwhen the terminal potential difference is zero.The current corresponding to zero value ofterminal potential difference is 0.28 A. This ismaximum value of current.

r=!=~il; r=5il.I 0.28

(ii)

(1)

lPO

1=11+12

E1 = 2E

11 11---./ININ'-(1 = 2r

A12 E2= E 12

---./ININ'-(1 = (

IIII Eeq I

0 II 0A ---./ININ'- B

(eq

For cell-I2E - VV =VA - VB =2E - II (2r) =)11 =--

2r

For cell-If, V = VA - VB = E - I2rE-V

=) 12=--r

... (ii)

... (ill)

.,' From Eqs. (ii) and (iii), substituting in Eq. (I),I=2E-V+E-V

2r r

On rearranging the term, we get

v=4:-IenBut for equivalent of combination,

V = Eeq - I (req)

On comparing,E = 4E r. = 2r

eq 3' eq 3

NOTE Two cells of emfs E1 and E2 and internal resistances (1

and (2 connected in parallel combination, thenequivalent emf

E - E1(2 + E2(1eq-

(1+ (2

Equivalent resistance, (eq = ~(1 + (2

In these type of questions, we have to look outthe connections of different cells, if the oppositeterminals of all the cells are connected, then theysupport each other, i.e. these individual emf's areadded up. If the same terminals of the cells are

22.

o Chopterwise CBSE Solved Papers PHYSICS

connected, then the equivalent emf is obtained bytaking the difference of emf's.Net emf of combination = EI - 2EI + 5EI = 4EINet resistance of current = r + 2 r + 3r + R = 6r + R

V.. Current, 1= - (from Ohm's law)R

I=~6r + R2E1

3;~ ..(1)

R-

23. V=(R!r)R=l+ErIR

=) with the increase of R, V increases

(1)

(112)

v

E

R- (1%)

24. Refer Ans. 24

(1)

R(O)Graph between terminalvoltage (V) and resistance R

(1%)

(112) One can determine the emf of cell by findingterminal potential difference when current Ibecomes zero. (1/2)

25. Refer Ans. 25.

I (A)Graph between terminalvoltage (V) and current (1)

When R = 4 n and I = 1 A .


We know that, terminal voltage, V = E - Ir.~ V = IR = 4 = E - Ir~ E - r = 4 ... (i) (1)

When R = 9 0 and I = 0.5A , thenV=IR=0.5x9=E-0.5r

~ E - O.5r = 4.5 ...(ii) (1)On solving Eqs. (i) and (ii). we get

r = 10 and E = 5V. (1)

26. Applying Kirchhoff's second law in mesh AFEBA

-lxI-lxI-6+9=0-2l+3=0

I=~A ... (i)2 (1/2)

1 t n EFr---~~~-r---------,D

1

HlR

4n

n ~vnA'-----j~

9V 3V W~Applying Kirchhoff's second law in mesh AFDCA-lx1-1xI-IxR-3+9=0

- 21 - IR + 6 = 021 + IR = 6

From Eqs. (i) and (ii), we get

(2X~) + ~ R = 6

R=20

1

c

(112)

... (ii)

(1/2)

For potential difference across A and D along AFD3 3VA - - x l - - x l = VD2 2

VA - VD = 3 V (1)

27. For BCD, equivalent resistanceRI = 50 + 50 = 100 (112)

Across BA, equivalent resistance R21 1 1 1-=-+-+-

R2 10 30 15

3+1+2=~=~30 30 5 (112)

~ R2 = 50Potential difference,

VBA = I X R2 = 0.2 x 5 (1)

VBA = 1 V ~ VAB = - 1 V (1)

1

28.11 R1 =4n A (13 + 14)

012 0 14 13

E = 10VR2 =15 n 0 R3 = 15n

R4 = 30n

CB

According to figure 150, 30 nand 150 are inparallel. their equivalent resistance (Reqlis

_1_ = ~ + ~ + ~= 2+ 1+ 2 = 2Req 15 30 15 30 301 1

Req 6

Req=60Now, Req = 6n and 40 are in series theirequivalent resistance R~ is

R~ = Req + 40 = 60 + 40 = 100By junction rule at node A

II = 12 + 13 + 14 ... (i) (1/2)

Applying Kirchhoff's second rule in(i) In meshADB,

-14 x 30+ 1512= 0

I2=2I4~I4=I22 (112)

(ii) In mesh BDC,

3014 -1513 = 0

13 = 214 ~ 14 = 132

In mesh ABE (containing battery), .J'

-411 -1512 + 10= 0411 + 1512= 10

In meshABCD,

-1512+ 1513 = 0

I • I 121=12+ 2 +-

2From Eq. (ii), we get

4GI2)+ 1512= 10I

10 2 2I2=-A=-A=I3 ~I2=I3=-A

25 5 5I4=I2=~A

2 55 5 211=-12 =-x-=IA2 2 5

rO

(ill) (1/2)

(iv)... (ii)

(112)

(1)

102

29. Kirchhoff's first rule or junction rule Thealgebraic sum of electric currents at any junctionof electric circuit is equal to zero i.e. the sum ofcurrent entering into a junction is equal to thesum of current leaving the junction:::) I.l = 0At junction 0, II + 12 = 13 + 14

X'o"

12 14

(1)

Kirchhoff's second rule In any closed mesh ofelectrical circuit, the algebraic sum of emfs of cellsand the product of currents and resistances isalways equal to zero.i.e. IE + r.IR = 0Kirchhoff's second law is a form of law ofconservation of energy. (1)

For given circuit,E1 = 2V '1 = 4 n

B11 012 E2 = 1 V '2 =3n

C

013 E3 = 4V '3 =2 n

A .--+---i

F+----i

E L...----if---M"N'---'D

At F, applying junction rule13=11+12

In mesh ABCFA,

-2- 411 + 3[2+ 1= 0411 - 3[2 =-1

In mesh FCDEF,

-1-312-213+4=0

312+213=3On solving, we get II ,12 and 13.

2II =-A,13712=-A139and 13 =-A13

... (i)

... (ii)

... (iii)(1)

IZI Chapterwise CSSE Solved Papers PHYSICS

30. For Kirchhoff's Rules, refer to Ans. 30.

The given diagram E - V1-6

P

(w.)

A B

13

cR = 12n

Applying Kirchhoff's second rule in loop ACBPA,

-1213 + 6 - 0.511 = 0511 + 12013 = 60 ... (i)

In loop ACBQA,

-1213+ 10 -12 x l = 01213+12=10 ... (ii) (112)

Also Kirchhoff's junction rule,11+ 12= 13 ... (iii) (112)

(Here, three equations are the expressions for11,12 andJj)On solving Eqs. (i), (ii) and (iii), we get

8411=--A37

:::) 12 = 106 A37

22:::) 13=-A37 (112)

31. For statement of Kirchhoff's rules, refer toAns.30. (1)

Applying Kirchhoff's second rule to the loop PRSP,I.E+ r.IR = 0

-13x20-12x200+ 5=041) + 4012 = 1 ... (i)

For loop PRQP,

- 2013 - 6011 + 4 = 0513+ 1511 =1 ... (ii)

Applying Kirchhoff's first rule at P

13 = II + 12 ... (iii) (1)

From Eqs. (i) and (iii). we have411 + 4412 = 1 ... (iv)


From Eqs. (ii) and (iii), we have2011 + 512 =1 ... (v)

On solving the above equations, we getI =.!.!..A = 11000 mA

3 172 172

I =~A=400°mA2 215 215

and I = ~A = 39000mAI 860 860 (1)

32. The high resistance voltmeter means that nocurrent will flow through it hence, there is nopotential difference across it. So, the readingshown by the high resistance voltmeter can betaken as the emf of the cell.The internal resistance of a cell depends on(i) the concentration of electrolyte and

(ii) distance between the two electrodes.(112 x 2=1)

The emf of cell (E) = 2.2 VThe terminal voltage across cell when 5 nresistance (R) connected across it (V) = 1.8 VLet internal resistance = r.: Internal resistance,

r=R(~-l).. r=5(2.2_I)=5XO.4=2=1On

1.8 1.8 1.8 9

~ r = 10 n9 ro

33. (i) Refer to Ans. 30. 1)(ii) Let 61 current be drawn from the cell. Since,

the paths AA', AD and AB are symmetrical.current through them is same.As per Kirchhoff's junction rule, the currentdistribution is shown in the figure. (1)

21 610'

211 B'

A'

21

61

(1)

103

34.

Let the equivalent resistance across thecombination be R.

E = VA - VB = (61) R~ 6IR = 10 (.: E = 10V) ... (i)Applying Kirchhoff's second rule in loop AA'B'C'A

- 2I x I - I x I - 2I xl + 10 = 0~ 51 = 10

1 =2ATotal current in the network = 61 = 6 x 2 = 12 AFrom Eq. (i),

61 R =10 ~ 6 x2x R = 10

R = 10 = 2 n ~ R = 2 n12 6 6 (1)

The current through 2n resistor is 0.9 A.Let II and 12 be the currents in two cellswith emf's. EI and E2 and internal resistances,rl and r2•

E" (,

A----< )----B

So, 1= II + 12Now, let V be the potential differencebetween the points, A and B. Since, thefirst cell is connected between the pointsA and B.

V = potential difference across first cell

V = EI - II rl or II = EI - Vrl (1)

Now, the second cell is also connected betweenthe points, A and B. So,

I2=E2-V

r2

Thus, substituting for II and 12

I=EI-V+E2-Vrl r2

or 1=(5..+ E2)_V(..!..+..!..)7j r2 rl r2

v=(Elli+E2rl)_I(_rlr2) ... (i)rJ + r2 rJ + r2

104

If E is effective emf and r, the effective internalresistance of the parallel combination of the twocells. then

v = E - Ir

Comparing Eqs. (i) and (ii), we get

(i) E = El r2 + E2 r1

r1 + r2

This is equivalent emf of the combination.(ii) r=~

r1 + r2

This is equivalent resistance of thecombination.

(ill) the potential difference between the points Aand Bis (1)

... (ii) (1)

v = E - Ir35. No current flows through 4 Q resistor as capacitor

offers infinite resistance in DC circuits.Also, 2Q and 3Q are in parallel combination

z x 3 6RAB = -- = -=1.2A

2+ 3 5Applying Kirchhoff's second rule in outer loop ABand cell.Let I current flow through outer loop in clockwisedirection.

-1.21 - 2.81 + 6= 04I = 6

1=~A2 (1'12)

:. Potential difference across AB3

VAB = 1RAB = - x 1.2 = 1.8V2

:. 3Q and 2Q are in parallel combination.:. Potential difference across 2Q resistor is 1.8 V.:. Current I' through 2Q resistor is given by

t: =.!:::= 1.8 = 0.9AR 2

t: = 0.9 A (1'12)

o Chnpterwise eBSE Solved Popers PHYSICS

36. For statement of Kirchhoff's laws Refer toAns.29. (2)

Let I current flows clockwise in the circuit.Applying Kirchhoff's voltage rule

- 2 - 1rl - 1r2 + 2 = 0 (112)

1rl + 1r2 = 01(rl + r2) = 0

• • 1j + r2 *- 0 ~ I = 0 (112)

Thus, no current flows through the circuit.

37. (i) Refer to Answer 30. (2)(ii) The circuit diagram can be redrawn as given

below.

B

E,r(Equivalent circuit)

r rRAC = -; ReB = -

2 21 1 1 1 3~ - =-+-+ -=-

RAB r r r r

RAB =!..3

Total resistance of circuit = r +!.. =~3 3

Current drawn from cell = _E_ = ~4r/3 4r

( )

22 3E 9E 2

Power consumed = I R = - r = --2 . r4 r 16 r

9E2

16 r (3)

[TOPIC 3] Electrical Devices and HeatingEffects of Current

3.1 PotentiometerIt is an electrical device which can

(i) measure the potential difference withgreater accuracy.

(ii) compare the emfs of two cells.(ill) measure the emf of a cell.(iv) be used to determine the internal

resistance of a primary cell.

Working PrincipleThe potentiometer works on the principlethat potential difference across any twopoints of uniform current carryingconductor is directly proportional to thelength between the two points i.e. V oc I

Potential GradientIt is the potential drop per unit length ofwire of potentiometer

i.e. K = ~, where V and I are potentialI

difference applied by driving cell and lengthof wire of potentiometer, respectively.

Application of PotentiometerMeasurement of Potential DifferenceUsing Potentiometers

+~~.E K

o 100I.I.I.I,I.I.I,I.I.I.I.I.!.I.I.I.I,I.!.I.I.I.I.I.I,I.I.!,I.I.I,I.,I

B

I--(

E1 K1If r is the resistance of potentiometer wire oflength L, then current through

. .. I Epotentiometer WIre IS =--R+r

Potential drop across potentiometer wire =Ir=(~)rPotential gradient of potentiometer wire R+ r

K = (~)!..~V = KI=(~)!..IR+r L R+r L

The potentiometer is a better device to measurepotential difference than a voltmeter as null pointmethod is used and hence, it can measure even the emfof cell but voltmeter cannot. It measures potentialdifference with greater accuracy.

Comparing EMF of Two CellsThe emfs of two primary cells can be compared using

. EI IIpotentiometer as - = -E2 12

where, II antll2 are the balancing lengths correspondingto cells of emf's EI and E2 respectively.

Circuit diagramFor comparing the ernfs of two primary cells.

V K Main-1 ~ ~ f--{. circuit

Af J' / -J-,,-R:!..h----I B

[j'E1 K1I-----{.

E2 K2I-----{

To Measure Internal Resistance of a CellThe internal resistance can be determined usingpotentiometer.

V K1111f--{·~Rh Main

circuitJ' J J"A~---~'-'-T-----~B

AS

106

If 11,12 are the balancing lengths when key KI isopened and closed respectively and resistance R isapplied in Safety Resistance Box (SRB), theninternal resistance of primary cell of emfs is given

by r=R(t-l)The potentiometer works only when

(i) the terminal voltage applied by driving cell isgreater than the emf of primary cell.

(ii) the positive terminals of driving cell andprimary cell are connected at the zero end ofpotentiometer wire.

Sensitivity of a PotentiometerIt refers to the capability of measuring very smallpotential difference and exhibit change inbalancing length even on very small change inpotential difference.The sensitivity of potentiometer can be increasedby increasing the number of wires ofpotentiometer and hence, decreasing the value ofpotential gradient.

3.2 Wheatstone BridgeIt is an arrangement of four resistance connectedto form the arms of quadrilateralABCD. A batterywith key and galvanometer are connected alongits two diagonals respectively.

B

oE'-------lll----{(. }-----'

K

where, P, Q = ratio arms,R =known resistance,S = unknown resistance

The bridge is said to be balanced, when(i) VB = VD

(ii) There is no flow of current throughgalvanometer,


The Wheatstone bridge is said to besensitive, if it gives ample deflection in thegalvanometer even on slight change ofresistance. For sensitivity of galvanometer,the magnitude of four resistances P, Q, R, Sshould be of same order.

3.3 Meter BridgeR s

Altr---~~~~---~I-(100-1)E KIIf--(.)--.IWW-

It is an electrical device used to determinethe resistance and hence, specific resistanceof material of given wire/conductor.It is based on the principle of balancedWheatstone bridge.For uniform wire,resistance of wire oc length of conductorAt balanced situation of bridge,

P R-=-Q S

Ror --=-100 -I S

~ s=COOI-I)XRwhere, 1 is the balancing length.

NOTE Meter bridge is also known as slide wire bridge.

3.4 Joule's Law of HeatingThe amount of heat produced in a current carryingconductor, H is given by

Hoc [2 Rt or H = [2 Rt

V2H = VIt or H=-t

RR (resistance)

-----IVV'N'-I (current)and t (time)


Electric PowerThe rate of consumption of electrical energy orproduction of heat energy is known as electricpower. SI unit of electric power is watt (W).

V2Electric power, P = [2R = -- = VI

R

• Unit in the commercial unit of electric energy,1unit = 1kWh = 3.6x 106 J

• If V, = specified voltage and W is wattage ofbulb or appliance, then resistance of bulb of

V2appliance _= _s_

W

If Va i' applied voltage, then actual powerv2 V2

consumed, Pa = ----"--= -1WR Vs

PREVIOUS YEARS'EXAMINA .ION QUESTIONSTOPIC 3o 1 Mark uuestlons

1. Nichrome and copper wires of samelength and same radius are connectedin series. Current I is passed throughthem. Which wire gets heated up more?Justify your answer. All India 2017

2. State the underlying principle of apotentiometer? Delhi 2014C

3. A heating element is marked 210 V,630 W. What is the value of the currentdrawn by the element when connectedto a 210 V DC source? Delhi 2013

4. A resistance R is connected across a cellof emf E and internal resistance r. Now, apotentiometer measures the potentialdifference between the terminals of thecells as V Write the expression for r interms of E, Vand R. Delhi 2011, 2010

107

5. In an experiment on meter bridge, if thebalancing length AC is X, what would beits value, when the radius of the meterbridge wire AB is doubled? Justify youranswer. All India 2011C

6. In a meter bridge, two unknown resistancesRand S when connected in the two gaps,give a null point at 40 em from one end .What is the ratio of Rand S? Deihl 2010

o 2 Marks Questions7. Use Kirchhoffs rules to obtain balance

conditions for the balance conditions in aWheatstone bridge. All India 2015

8. (i) State the principle of working of apotentiometer.

(ii) In the following potentiometer, circuitAB is a uniform wire of length 1m andresistance 10 n. Calculate the potentialgradient along the wire and balancelength AO(l).

I 2V 15Q

Ar-------~------~B1.2 Q I

O.3Q Delhi 2015C

9. A potentiometer wire of length 1 m has aresistance of 10 n. Determine the emf of theprimary cell which gives a balance point at40 cm. Delhi 2014

Or A potentiometer wire of length 1 m has aresistance of 5 n. It is connected to a 8 Vbattery in series with a resistance of 15 n.Determine the emf of the primary cell whichgives a balance point at 60 cm. Delhi 2014

Or A potentiometer wire of length 1.0 m has aresistance of 15 n. It is connected to a 5 Vbattery in series with a resistance of 5 n.Determine the emf of the primary cell whichgives a balance point at 60 cm.Delhi 2014; HOTS;Foreign 2012

108

10. An ammeter of resistance 0.80 n canmeasure current up to 1.0 A.

(i) What must be the value of shuntresistance to enable the ammeter tomeasure current up to 5.0 A ?

(ii) What is the combined resistance ofthe ammeter and the shunt? Delhi 2013

11. Describe briefly with the help of a circuitdiagram, how a potentiometer is used todetermine the internal resistance of a cell.All India 2013

12. Two studentsXandYperform an

. A~------~r-------~Bexperiment onpotentiometerseparatelyusing thecircuit givenbelow.

, Keeping other parameters unchanged,how will the position of the null point beaffected, if

(i) X increases the value of resistance Rin the set up by keeping the key Klclosed and the key K z open?

(ii) Y decreases the value of resistance Sin the set up, while the key Kzremains open and then Kl closed?Justify your answer. Delhi 2017

13. In the meter bridge experiment, balancepoint was observed at J with AJ = l.

(i) The values of R and X were doubledand then interchanged. What wouldbe the new position of balance point?

(ii) If the galvanometer and battery areinterchanged at the balancedposition, how will the balance pointget affected?

R X

G

A <-----------''-------------Ia

All India 2011


14. For the circuit diagram of a Wheatstonebridge shown in the figure, use Kirchhoffslaws to obtain its balance condition.

+ -Delhi 2009

15. Obtain the formula for the power loss (i.e.power dissipated) in a conductor ofresistance R, carrying a current. Delhi 2009C

o 3 Marks Questions16. A resistance of R

draws currentfrom apotentiometer.The potentiometerwire AB, has atotal resistance of Ro. A voltage V issupplied to the potentiometer. Derive anexpression for the voltage across R, whenthe sliding contact is in the middle ofpotentiometer wire. Delhi 2017

17. (i) The potential difference appliedacross a given resistor is altered, sothat the heat produced per secondincreases by a factor of 9. By whatfactor does the applied potentialdifference change?

(ii) In the figure shown, an ammeter Aand a resistor of 4 n are connected tothe terminals of the source. The emfof the source is 12 V having aninternal resistance of 2 n. Calculatethe voltmeter and ammeter readings.

v

Ar---~NV~----~a

R

'VI/IM.-----------'R=4 n All India 2017


18. (i) Write the principle of working of ametre bridge.

(ii) In a metre bridge, the balance point isfound at a distance '1 with resistance Rand S as shown in the figure.

R s

G

A~------~~--------~B'-------11--< }-----'

An unknown resistance X is nowconnected in parallel to the resistanceS and the balance point is found at adistance 12, Obtain a formula for X interms of 11,12and S. All Indio 2017

19. Answer the following:(i) Why are the connections between the

resistor in a meter bridge made ofthick copper strips?

(ii) Why is it generally preferred to obtainthe balance point in the middle of themeter bridge wire?

(iii) Which material is used for the meterbridge wire and why? All India 2014

20. (i) State the underlying principle of apotentiometer. Why is it necessary to(a) use a long wire,(b) have uniform area of cross-section

of the wire and(c) use a driving cell whose emf is

taken to be greater than the emfs ofthe primary cells?

(ii) In a potentiometer experiment, if thearea of the cross-section of the wireincreases uniformly from one end tothe other, draw a graph showing howpotential gradient would vary as thelength of the wire increases from oneend. All India 2014C

21. In the figure, a long uniform potentiometerwire AB is having a constant potentialgradient along its length. The null pointsfor the two primary cells of emfs El and E2

109

E2 connected in the manner shown, areobtained at a distance of 120 cm and300 cm from the end AFind

(i) Elf E2 and(ii) position of null point for the cell E1.

How is the sensitivity of a potentiometerincreased? Foreign 2014; Deihl 2012

22. Define the current sensitivity of agalvanometer. Write its SI unit.Figure shows two circuits each having agalvanometer and a battery of 3 V.When the galvanometer in eacharrangement do not show any deflection,obtain the ratio Rtf R2•

L- -< G \- .J

1.20n

All India 2013

23. State the underlying principle of apotentiometer. Write two factors onwhich the sensitivity of a potentiometerdepends. In the potentiometer circuitshown in the figure, the balance point isat X. State, giving reason, how thebalance point is shifted when

200n

3.0V

5V 60n

Q4V

(i) resistance R is increased(ii) resistance S is increased, keeping R

constant? All India 2013C

110

24. With the help of circuit diagram, explainhow a potentiometer can be used tocompare emf of two primary cells?Deihl 2011

25. Two heating elements of resistances ~and R2 when operated at a constantsupply of voltage V, consume powers P1and P2, respectively. Deduce theexpressions for the power of theircombination when they are in turn,connected in

(i) series and(ii) parallel across their same voltage

supply. All India 2011

26. In a meter bridge, the null point is foundat a distance of 40 em from A. If aresistance of 12n is connected in parallelwith S, then null point occurs at 50.0 emfrom A. Determine the values of Rand S.

120R

HOTS;Oalhi 2010

27. Draw the circuit diagram of apotentiometer which can be used todetermine the internal resistance r of agiven cell of emf E. Explain briefly howthe internal resistance of the cell isdetermined? Delhi 2010

28. In a meter bridge, the null point is foundat a distance of 60 em from A. If aresistance of 5 n is connected in serieswith S, then null point occurs at 50.0 emfrom A. Determine the values of Rand S.

R S

A B!I'iil""""""""'I'I',""""""""""""'"j11'1'1'I'1

Oalhi 2010

o Chopterwise CSSE Solved Papers PHYSICS

29. In a meter bridge, the null point is found at adistance of 11 em from A. If a resistance of X isconnected in parallel with S, then null pointoccurs at a distance 12 em from A. Obtain theformula for X in terms of 11' 12 and S.

X

R S

A~II~ij~'Ii~'i':;;;'ii:;;;'il;;;:'i'~I'~iij~'1i~I":;;;'Ii;;;:I";'I~'1i~I":;;;il':;;;1'1;;;:'li~l'~jll:;;;lji:;;;l'l~il~B

Oeihi 2010

30. A circuit using a potentiometer andbattery of negligible internal resistance isset up as shown to develop a constantpotential gradient along the wire AB. Twocells of emfs El and E2 are connected inseries as shown in combinations (1) and(2). The balance points are obtainedrespectively at 400 cm and 240 em fromthe point A. Find

(i) E,. / E2•

(ii) balancing length for the cell E,. only.III----( )----,

A B

I CD 'I1 E2 ,,I@,I

,0',

E2 E1 Delhi 2009

31. X Y

The figure shows experimental set up of ameter bridge. When the two unknownresistances X and Yare inserted, the nullpoint D is obtained ;10 cm from the end A.


32.

When a resistance of 10 n is connected inseries with X, the null point shifts by10cm.Find the position of the null point when theIon resistance is instead connected inseries with resistance Y. Determine thevalues of the resistances X and Y. Deihl 2009

(i) State the principle of working of ameter bridge.

(ii) In a meter bridge balance point isfound at a distance 11 with resistancesRand S as shown in the figure.When an unknown resistance X isconnected in parallel with theresistance S, the balance point shiftsto a distance 12, Find expression for Xin terms of 11, 12 ,and S. All Indio 2009

XR ,"""I/IN\, \

A~------~--------------~Bl...-_--j 1f-----1 }----------'+ -

33. (i) In the circuit diagram given below ABis a uniform wire of resistance15 n and length 1 m is connected to acell E1 of emf 2V and negligibleinternal resistance and a resistance R.The balance point with another cellE2 of emf 75 mV is found at 30 emfrom end A. Calculate the value of R.

E1 R

A~----rX~------~B

G(ii) Why is potentiometer preferred over

a voltmeter for comparison of emf ofcells?

(iii) Draw a circuit diagram to determineinternal resistance of a cell in thelaboratory. Foreign 2016

111

34. (i) State the working principle of apotentiometer. With the help of thecircuit diagram, explain how apotentiometer is used to compare theemfs of two primary cells. Obtain therequired expression used forcomparing the emfs.

(ii) Write two possible causes for onesided deflection in a potentiometerexperiment. Deihl 2013

o 5 Marks Questions35. (i) State Kirchhoffs rules for an electric

network. Using Kirchhoffs rules,obtain the balance condition in termsof the resistances of four arms ofWheatstone bridge.

(ii) In the meter bridge experimental setup, shown in the figure, the null pointD is obtained at a distance of 40 emfrom end A of the meter bridge wire.

If a resistance of IOn is connected inseries with~, null point is obtained atAD = 60 em. Calculate the values of~ and s;

BA1I1111,IIII,IIII,IIII,IIIIIIIIII"II,lllIllllljllIl,111I111111I

E'-------0.1- ----{ }-----' Delhi 2013

36. (i) Use Kirchhoffsrules to obtainthe balancecondition in a AWheatstonebridge.

(ii) Calculate thevalue of R in thebalance l...-----~Vf------...Jcondition of theWheatstone bridge, if the carbonresistor connected across the arm CDhas the colour sequence red, red andorange, as shown in the figure.

B

c

o

112

(iii) If now the resistance of the arms BCand CD are interchanged, to obtainthe balance condition, another carbonresistor is connected in place or R.What would now be sequence ofcolour bands of the carbon resistor?

Delhi 2012

37. (i) State the working principle of apotentiometer. Draw a circuit diagramto compare emf of two primary cells.Derive the formula used.

(ii) Which material is used forpotentiometer wire and why?

(iii) How can the sensitivity of apotentiometer be increased? Delhi 2011

38. (i) State with the help of a circuitdiagram, the working principle of ameter bridge. Obtain the expressionused for determining the unknownresistance.

(ii) What happens if the galvanometerand cell are interchanged at thebalance point of the bridge?

(iii) Why is it considered important toobtain the balance point near themid-point of the wire? Delhi 2011e

o Explanations1. For same length and same radius, resistance of

wireRocp (p: resistivity)

As A,lChrome > Pcopper

Hence, resistance of nichrome section is more.In series, same current flows through both sectionsand heat produced = [2Rt. So, more heat isproduced in nichrome section of wire. (1)

2. The potentiometer works on the principle thatpotential difference across any two points of uniformcurrent carrying conductor is directly proportional tothe length between those two points. (1)

3. Given that P = 630 W and V = 210 V.

In DC source P = VI.P 630Therefore, I = - = - = 3 A.V 210

4. Internal resistance, r = R( % - 1)

where, signs are as usual.

o Chopterwise CBSE Solved Popers PHYSICS

5. The balancing length continue to be X even ondoubling the radius of meter bridge wire as itdoes not affect the ratio of length of two parts ofmeter bridge wire. (1)

NOTE Resistance of wire =(%}For uniform wire, (%) is constant even on doubling the

radius of meter bridge wire.:. Resistance of wire oc l.

6.. : Null point is obtained at 40 cm from one end/ = 40cm,

100 -/ = 60crn.For meter bridge ratio of unknown resistances

R / 40 2-=---=-=-S (l00 - /) 60 3

=> R : S = 2 : 3 (1)

7. Applying Kirchhoff's loop law to close loop ABDA,we get II R - IsG - 12 P = 0 ... (i)Consider the diagram

8I2-Ig

19 Q

oI I

Here, G is the resistance of the galvanometer.Applying Kirchhoff's loop law in the closed loopBDCB, we get

IsG + (II + Is)S - (12 - Is) Q = 0 ...(ii)When the Wheatstone bridge is balanced, nocurrent flows through the galvanometer,i.e. Is = 0 (1)

:. From Eq. (i), we getI I I I II =!:..IR- 2P=0=> IR= 2P => ... (iii)

12 R

(1)

Similarly, from Eq. (ii), we getliS - 12 Q= 0

II S = 12 Q => II = gI S

From Eqs. (ill) and (iv). we g~t!:..=g => !:..=!!.R S Q S

This is the required balance condition in aWheatstone bridge arrangement.

... (iv)

(1) (1)


8. (i) Refer to Ans. 2.(ii) Here AB=lm,R AB =10n,

Potential gradient, k =?,AO=I=?Current passing through AB,

1= __ 2_ = _2_ = 2A15+RAB 15+10 25

2 4VAB = IxRAB = -xl0 = - V25 5

k= VAB =~vm-IAB 5

Current in the external circuit,I' = ~ = 1.5= 1A

1.2+0.3 1.5

For no deflection in galvanometer,Potential difference across AO=1.5-1.2I'~ k(1)=1.5-1.2xI'

~ ~l = 0.35

1=0.3x5 =0.375m4

1=37.5cm

or

9. Given, length of wire, I = 1 m = 100 cm

Resistance, R = 10 nEmf of a battery, EI = 6 V,

RI = 5n andx=40cm

E 6 6:. Current, I = __ I -=--=- AR+RI 10+515

6 60VAB =IR=-x 10=-=4V

15 15

:.Emf of the primary cell = VAB X xI

=~X40=1.6V100

Or (Ans. l.2V).

Or (Ans. 2.25 V).

10. (i) To measure current up to 5A, the shunt Sshould have a value, such that for 5A inputcurrent through system, 4A should passthrough shunt S and 1 A should pass throughgiven ammeter.

1xRA =4S1x 0.8 = 4S ~ S = 0.2 n

Thus, the shunt resistance is 0.2 n. (1)(ii) Combined resistance of the ammeter and the

shunt,R = 0.8S = 0.8x 0.2 = 016 = 016n.

0.5+ S 0.8+ 0.2 1 (1)

113

(1) 11. Measurement of internal resistance of a cell usingpotentiometer.

(1)

(1)

The cell of emf, E (internal resistance r) is connectedacross a resistance box (R) through key K 2'

E = <jl4 ... (i)When K2 is open balance length is obtained atlength ANI = ~.. V=<I>l.JFrom Eqs. (i) and (ii), we get

!i=iV l.J

E = l(r + R) ~ V = IRE r+ R----V R

From Eqs. (ill) and (iv), we getR+ r =i

R l.J

!i=iV l.J

r=R(%-l)

r = R(i -1)

... (ill)

... (iv)

(112)

(112)

(1)

(2)

(2)

We known ~,l.J and E, so we can calculate r. (1)

12. When KI is closed and K2 is open, then only thecell connected in upper part branch will work.When K2 is closed and KI is open, then only thecell connected in lower branch will work.(i) KI ~ closed, K2 ~ open

E K,+~.

114

Suppose null point occurs at J.Apply KVL in smaller loop,

E-IR=O

where, R = resistanceE=IR=>I=!!..

RAs, X increases the value of resistance R. So,current in the circuit (wire) decreases. Hence, Rwill be increased. Then I will decrease.We can say, as X increases the value of R nullpoint decrease. (1)

(ii) K2 ~ open, K) ~ closed.Then the circuit will be same as shown earlier.We see that resistance S is not involved in thecircuit because K 2 is open.

... (i)

JA.---------~.-----------~B

+

K2

So, from Eq. (i). we getE =RI

I .s.R

Here, R does not depend on the value ofresistance S.So, R null point is not affected by decreasing thevalue of resistance S. (1)

13. (i) The balancing condition states thatR /- - ------X (l00 - /)

X 100-/- ------R /

When both X and R are doubled, then2X X 100-/------2R R /

Balancing length would be at (l00 - /) ern.(1)

On changing the position of galvanometer andbattery, the meter bridge continue to bebalanced and hence no change occur in thebalance point. (1)

(ii)

o ehopterwise eBSE Solved Papers PHYSICS

14. In balanced Wheatstone bridge if current flowthrough galvanometer, that means while applyingKirchhoff's law, we can neglect this path.

No current flows through the galvanometer Gwhen circuit is balanced.

B[, I,

I

o'- .L'-I, _

EOn distributing currents as per Kirchhoff's firstrule.Applying Kirchhoff's second rule(i) In mesh ABDA,

.. -I) R) + (I - Id R4 = 0=> I) R) = (I - Id R4 ... (i)

(ii) In mesh BeDB,-I) R2+ (I - Id R3 = 0

=> I)R2=(I-IJ)R3 ... (ii)

On dividing Eq. (i) by Eq. (ii). we getI) RJ _ (I - I)) R4.

I) R2 - (I - I)) R3'

(1)

15.

!i = R4R2 R3

This is necessary and required balanced conditionof balanced Wheatstone bridge. (1)

Consider a conductor MN having resistance R.M R N~

The potentials of the two terminals are suppose VM

and VN•

Such that, VM - VN = V

At any time interval fll,

conductor will be

(As, VM > VN)

current through the

1= &jfll (1)

where, Sq = charge drifted through the conductor.The electrical potential energies of the charge&j at M and N are flU M = &jV M and flU N = &jVN,

respectively.:. Change in potential energy

flU = tlUN - flU M


As, loss of potential energy = gain in kinetic energy~ ilK = - flU = [Vflt (1)

:. Energy dissipated per unit time (called powerloss) is

p = IVflt = VI = I2R (.: V = IR)M

16. Total resistance of the circuit = Ro + Req2

1 1 2 Ro+2R RRo- =- +- = --- ~ Req=---Req R Ro RRo Ro + 2R (1)

Total resistance= Ro +~= ~ + 2RRo + 2RRo = ~ + 4RRo

2 Ro + 2R ~Ro + 2R) ~Ro + 2R)

Current in the circuit = V = 2V (Ro + 2R)~ + 4RRo ~ + 4RRo

~Ro + 2R) (1)

Current in R 2V(Ro + 2R) x Ro /2(~ + 4RRo) (R + Ro /2)

= V(Ro + 2R) x 2Ro 2VRo

(~ + 4RRo) (2R + Ro) (~ + 4RRo)

Potential difference acrossR = 2V RoR 2VR

(~ + 4RRo) (Ro + 4R)V2

17. (i) Heat produced per second = 12R =-R

So, when voltage is made three times, heatproduced increase nine times for same R. (1%)

(ii) Current in the circuit is I = ~ = ~ = 2AR+r 4+2

Also, terminal voltage across the cell,V = E - Ir = 12 - 2 x 2= 8 VSo, ammeter reading = 2Aand voltmeter reading = 8 V (1'12)

18. (i) A metre bridge is based upon principle ofWheatstone's bridge.

Under balance (I9 = 0) condition, ~ = ~Q 5

115

In case of a metre bridge, resistance R and 5 aretaken in form of a wire. (1)

+ -

(1)

In balance condition, ~ = __ 5_11 100 - 11

(ii) In given metre bridge, initiallyR SII 100 -II

When a resistance X is placed in parallel with

5 h . . 5Xr t en net resistance m gap = --

5+ XSo, in balance (with 5 and X are in parallel), (1)

R (S¥X)12 100 - 12

Substitute the value of R from Eq. (i) toEq. (ii).we get

". (i)

". (ii)

II X12(100 - II) (S + X)(100 - 12)

~ 5 + X= x{/2(100 -/d}11(100 - /2)

~ !... + 1 = /2(100 - 11)X 11(100 - 12)

!... = /2(100 - /1) - /1 (100 - /2)

X 11(100 - 12)

X= 5/1(100- /2) n100 (12 - /d

~(1)

19. (i) The connections between the resistors in ameter bridge are made of thick copper stripsbecause of their negligible resistance. (1)

(ii) It is generally preferred to obtain the balancepoint in the middle of the meter bridge wirebecause meter bridge is most sensitive whenall four resistances are of same order. (1)

(iii) Alloy, manganin or constantan are used formaking meter bridge wire due to lowtemperature coefficient of resistance and highresistivity. (1J

116 IZI Chapterwise CBSE Solved Papers PHYSICS

20. (i) Principle of PotentiometerRefer to Ans. 2.(a) We use a long wire to have a lower value

of potential gradient (i.e a lower "leastcount" or greater sensitivity of thepotentiometer.

(b) The area of cross-section has to beuniform to get a 'uniform wire' as per theprinciple of the potentiometer.

(c) The emf of the driving cell has to begreater than the emf of the primary cellsas otherwise, no balance point would beobtained. (2)

P ial di K V IR IPotenti gra lent, = - = - = -L L A

:. The required graph is as shown below

(il)

AreaA _

21. (i) Let potential gradient be K.E1-E2=Kx120 ... (i)

(cells are connected in opposite order)EI + E2 = K x 300 ... (ii)

(cells are connected in supporting order)

(il)

EI + E2 = K x 300 = ~EI -E2 K x120 2

Now, apply componendo and dividendo(EI + E2) + (EI - E2) = 5 + 2(EI +E2)-(E1 -E2) 5-2

5.=2E2 3

.. EI _ 7 E 3 E--- 2=- 1

. E2 3' 7

From Eq. (i), we get3 4

EI--EI =Kx120, -EI =Kx1207 7

7EI = K x 120 x-

4

(112)

=> EI = K x 210Null point for EI is obtained at 210 cm.

(1)

(1)

(1)

22.

The sensitivity of potentiometer be increased byincreasing the length of wire. (112)

Current sensitivity of a galvanometer is defined asthe deflection produed in galvanometer per unitcurrent flowing through it. Its SI unit isradian/ampere.

3.0V (1)

For balanced Wheatstone bridge, there will be nodeflection in the galvanometer.

4 6 4x 9-=-=>RI =--=60RI 9 6

12Q

3.0V (1)

For the equivalent circuit, when the Wheatstonebridge is balanced, there will be no deflection in thegalvanometer.

12 68 R2

R2 = 6 x 8 = 4 012

RI =~=~R2 4 2 (1)

23. For principle, Refer to Ans. 2. (1)

The two factors on which the sensitivity of apotentiometer depends are(a) the value of potential gradient (K). (1/2)(b) by increasing the length of potentiometer wire.

From the circuit diagram,(i) if R is increased, the current through the

potentiometer wire will decrease.


24.

Due to it, the potential gradient ofpotentiometer wire will also decrease. Thus,the position of J will shift towards B. (1)

(ii) if S is increased, keeping R constant, theposition of J will shift towards A. (112)

The required circuit diagram is shown in thefigure below.

400

J 300

70 80 90

The main circuit comprises of battery of emf E,key (K) and rheostat (Rh ). The auxiliary circuitcomprises of two primary cells of emfs EI and E2,

galvanometer, jockey and resistance box (RB) toprevent large current flowing through thegalvanometer.When key KI is closed and K2 kept open, the cell,EI comes into action. The jockey J is moved onthe wire AB till null point is obtained ingalvanometer. Let null point is obtained at lengthII' then emf of first cell is given by

EI = kl, ... (i) (1)

where, k is the potential gradient along the wire ABdue to battery E.Now, key K2 is closed and KI kept open and nullpoint is obtained at length 12, then

E2= kl2

~= kll =iE2 kl2 12

EI =iE2 12

... (ii) (1)

Therefore,

NOTE The null point is obtained only when(i) emf of battery E must be greater than emfs of two

primary cells E1 and E2 each.

(ii) all the positive terminals of cells and battery must beconnected at the same point.

(1)

117

25. To deduce the expression for the power of thecombination, first find the equivalent resistance ofthe combination in the given conditions.

V2 v2~=- =} RI =-

RI ~

v2 v2P2=- =}R2=-

R2 P2 (1/2x2=1)

(i) In series combination,V2 V2

Rs = RI +R2 = -+ -~ P2

R. = RI + R2 = V2(~+ ~)= V2(~ + P2)~ P2 ~P2

Now, let the power of heating element in seriescombination be Ps'

V2 2.. P = = V = ~ P2

s RI+R2 V2(~+P2) ~+P2PI P2

p = ~ P2

s ~ + P2

(ii) In parallel combination,~ = ~ + ~ = _1_ + _1_ =!L+ P2

R" RI R2 V2 V2 V2 V2

~ P2

1 1-=-(~+P2)Rp V2

Now, power consumption incombination

(1)

parallel

Pp = ~ = V2( ~)

Pp = V2

[ :2(~+ P2)]

Pp=~+P2 (1)(1)

26. In case of meter bridge at null point condition, thebridge is balanced, i.e. we can apply the conditionof balanced Wheatstone bridge.

Applying the condition of balanced Wheatstonebridge,

R I 40402---- ----S 100 - I 100 - 40 60 3R 2S 3

The equivalent resistance of 12 nand S n inall I· 12S r.par e IS-- ••.

12+ S

... (i)(1)

(112)

118

Again, applying the conditionR _ 50_

1( 12S ) - 50 -12+ S

R=.~12+ S

(112)

... (ii)=>

From Eqs. (i) and (ii), we get

~S=~3 12+ S

12 + S = 18 => S = 6il

R = ~S = ~X6 = 4il3 3

R=4il ( 1/2 x 2 = 1)

(3)27.28.

Refer to Ans. 10.The condition of balanced meter bridge

~ = 60 = 60 = ~S 100 - 60 40 2R 3S 2

... (i)(1)

Again, applying the condition.when Sand 5il areconnected in series

.s.: 50 =>~=lS+ 5 50 S+ 5

.,,(ii)(1)

From Eqs. (i) and (Ii). we get

~S = S + 5 =>~S- S = 52 2S=IOil => R=~S=~XIO=15il

2 2R = 15 il, S= 10 il

29. Initially, for balanced Wheatstone bridge,

~ = __ /1_ =>R= __ /I_S .,,(i)S I00 - II 100 - II

When X is connected in parallel with S, then

[ S~ ]- (l0:~ 12)

S+ X

=> .e: = (100 -/2)RS+ X 12

=(100-/2) X (_/I_)S [from Eq. (i)l12 100- II

S:X -CJ(~~~=::)

o ehapterwise eSSE Solved Papers PHYSICS

S+ X = (!1.)(~)X II 100-/2

~+ I = 12(100 -It!X II (l00-/2)

~ = !1.(100 - II )-1X II 100- 12

~ = 100(12 -II) => X = II (100 - 12) SX II (100 - 12) 100 (/2 - II) (1)

30. (i) In combination 1, net emf of combination isEI + E2, whereas for combination 2 net emf isE2 -E/.

(1)

EI + E2 = KII

where, K = potential gradientII = 400 em

For combination 2,E2-EI=KI2

12= 240 emEI + E2 = KII = 400 = ~E2- EI KI2 240 3

EI + E2 = ~E2- E/ 3

Applying componendo and dividendo theorem,we get

(EI + E2) + (E2 - EI) = 5+ 3(EI + E2) - (E2 - EI ) 5 - 3

E2=~ => ~=~EI 2 E2 4

~= ~ => E2= 4EIE2 4

=> EI+E2=Kx400 => 5E=Kx400

K=~=~400 80

Now, let balancing length for EI is IIE

:. EI =K/I => E= - X II => II = 80 em80 (1)

Applying the condition of balanced Wheatstonebridge!.. = _1_, where I is the balancing

Y 100- Ilength from end A.

Initially, I = 40 ernX 40402 ----Y 100 - 40 60 3

X=~Y3

.,,(i)

where,". (ii)

(1/2)

(1)

(ii) .: (say)

(1/2)

(1)

31.

(1)

=>

" .(i)(1/2)


When 10 0 resistance connected in series with X,null points shift to 40 + 10 = SOern.oo X+10=SO=1

Y SO=> X+10=Y=> Y-X=10 ... (il)

(112)From Eqs. (i) and (Ii), we get

!.= 10 0 => Y = 3003X= 200 (1)

Now, 100 resistance connected in series with Yand let null point is obtained at length / em.

~=_I-Y+ 10 100-/

~ = _._1- (.,'X = 200, Y = 300)30+10100-1

1 /----2 100-1

100-/=2131= 100

1 = 100 cm = 33.33 ern3 (1)

So, null point is obtained at length 33.33 cm.32. (i) Meter bridge works on the principle of

balanced Wheatstone bridge.In balanced Wheatstone bridge,(a) no current flow through the

galvanometer.(b) VB = Vo

(c) !:..=.!iQ s (1Va)

where, P,Qare ratio arms,R = known resistance andS = unknown resistance.

Io

E(ii) Refer to Ans. 34.

119

33. As per the figure, total current through the wireAB is given by 1=~=_2_

R+r R+1S

The potential gradient of the wire is given byK =lx22.=_2_ X22.

100 R+1S 100

As, the balance point with cell E2 of emf 7S mV isfound at 30 em from end A

_2_xO.lSX 30=7Sx10-)R+1S

( 2 )XO.lSX30)-lS=R => R=10S07SxlO- (1'1a)

(ii) Potentiometer is preferred over a voltmeter forcomparison of emf of cells because at null point,it does not draw any current from the cell andthus there is no potential drop due to theinternal resistance of the cell. It measures thepotential difference in an open circuit which isequal to the actual emf of the cell. (1'Ia)

34. (i) Working principle of potentiometerRefer Ans. 24For derivation, Refer to Ans. 31. (1)

(ii) (a) The emf of the cell connected in maincircuit may not be more than the emf ofthe primary cells whose emfs are to becompared. (1)

(b) The positive ends of all cells are notconnected to the same end of the wire. (1)

35. (i) Refer to Ans. 34 of Topic 2.Wheatstone Bridge (1)The Wheatstone bridge is an arrangement offour resistances as shown in the followingfigure.

B

(1Va)

ERj, R2, R) and R4 are the four resistances.Galvanometer (G) has a current 19 flowingthrough it at balanced condition,

19 = 0Applying junction rule at B,oo 12 = 14

(1)

120

Applying junction rule at D,

.. II =13Applying loop rule to closed loop ADBA.

- II RI + 0 + 12R2 = 0!.L = R212 RI

... (i)

'I

Applying loop rule to closed loop CBDC,

12R4 + 0-IIR3 = 0

13=11 ~ 14=12

!.L = R412 R3

From Eqs. (i) and (ii),R2 = R4RI R3

This is the required balanced condition ofWheatstone bridge.

(ii) Considering both the situations and writingthem in the form of equations.Let R' be the resistance per unit length of thepotential meter wire

RI = R' x 40 _ 40 _ 3R2 R' (100- 40) 60 3

RI + 10 = R' x 60 = 60 = ~R2 R' (100- 60) 40 2

RI =3R2 3

RI + 10 = ~R2 2

Putting the value of RI from Eq. (i) andsubstituting in Eq. (ii}, we get

2103- + - = - ~ R2 = 1203 R2 2

Recalling Eq. (i) again,

RI =3 ~ RI = 8012 3

... (ii)

... (i)

... (ii)


36. (i) The balance condition is!:..=~~ !:..=gQ S R S (1)

(ii) Let a carbon resistor S is given to the bridge

~ ~=1S

~ R = S = 22 X 103 0 (1)

(ill) After interchanging the resistances thebalanced bridge would be

2R 2 x 103 1X = 2 x 22 X 103 = "2

~ X = 4R = 4 x 22 X103 = 88kO (1)

Thus, equivalent resistances of Wheatstonebridge

1 1 1 3-=-+-=- ~ Req=2RReq 3R 6R 6R

:. Current through it I = !. x ~ = ~ A3 2R 6R (1)

37. (i) For working principle of potentiometerRefer to Ans. 24. (1'/.)For circuit diagram to compare emf oftwo cells Refer to Ans. 31. (1'/.)

(ii) Constantan or manganin (alloy) as theyhave low temperature coefficient ofresistance.

(ill) The sensitivity of potentiometer can beincreased by increasing the number of wiresof potentiometer and hence, decreasing thevalue of potential gradient. (1)

38. (i) Refer to Ans. 33. (3)(ii) Refer to Ans. 13 (ii). (1)

(ill) It is because of the fact that meter bridge ismost sensitive when null point occur nearthe mid-point of wire and all the fourresistances are of same order. (1)

(1)

(1)

(3)

Value Based Questions (From Complete Chapter)o 4 Marks Questions

1. Ramaniamma was a childless widow. She ranher life only by the pension for the Sr. citizensfrom the Government. When she switches onone bulb in her house, all the other appliancesget switched off. She could not even spend foran electrician. Sujatha living nearby decided

to do something for her. She referred toPhysics books and learnt that the seriescombination for the household connectionshould be the reason. She called anelectrician and had the circuit changed toparallel combination. The problem wassolved and Ramaniamma was happy. She


thanked Sujatha for her help to solve theproblem.Read the above passage and answer thefollowing questions.

(i) What are the values possessed bySujatha?

(ii) Why for household a parallelcombination used? Give twoadvantages.

Ans. (i) Care for elderly people, empathy, willingnessto gain knowledge. [1%]

(ii) A parallel circuit can run several devices usingthe full voltage. [1]The two advantages are as follows:(a) If one apppliance fails, then other are

running normally.(b) If one of the device shorts, the other

devices will receive no voltage preventingoverload damage. [1%]

2. It is desired to supply a current of 2 Athrough a resistance of 10 Q. As many as20 cells are provided, each of emf 2 V andinternal resistance 0.5 Q. Two friendsShikhaj and Sanjeev try their hand on theproblem. Shikhaj succeeds but Sanjeevfails.Read the above passage and answer thefollowing questions.

(i) Justify the set up of Shikhaj?(ii) What might have gone wrong with

Sanjeev, when he gets 1.4 A currentin the external load?

(iii) What are the basic values shown byShikhaj and Sanjeev in their work?

Ans. (i) Here, n=20, E=2V .,=0.5 n,R=10 nIf all the 20 cells are correctly connected inseries to the external load of resistance R. thencurrent in load is

I=~R+ nr

20x2 = 2A10+ 20x 0.5

It means Shikhaj's set up is correct. [1]

(ii) If one cell is wrongly connected in seriesarrangement of cells, it then reduces the totalemf of the set up by an amount equals to twotimes the emf of each cell.

121

Let m cells be connected wrongly by Sanjeev inseries of total n cells, then he got the current1.4 A in the external load of resistanceR(= IOn). Therefore,

I = 1.4= (n - 2m) E (20 - 2m) x 2(R+nr) IO+20xO.5

On solving, we get m = 3. It means three cellsare connected wrongly by Sanjeev. [1%]

(iii) Shikhaj has proper knowledge and carefulhandling of the apparatus. That is why, he gotthe required results from his setup. Sanjeev'sknowledge is incomplete and his handling iscareless. That is why he could not get therequired results. Hence, to get proper results,one has to be extra careful and should plug allloopholes. [1%]

3. Vishwajeet purchased cells for histransistor. He felt that cells are notworking properly. He wanted to checktheir emf. So, he took the cells to thephysics lab and with the help ofpotentiometer found their emf. To hissurprise, emf was less than the valueclaimed by the manufacturer. He lodgedthe complaint with consumer forum andreceived the deserving response:Read the above passage and answer thefollowing questions:

(i) What values are displayed byVishwajeet?

(ii) Why do you think Vishwajeet usedpotentiometer instead of voltmeter tofind out emf of the cell? For moreprecise measurement, the potentialgradient of the potentiometer shouldbe high or low?

Ans. (i) Values displayed by Vishwajeet are as follow;(a) general awareness(b) presence of mind(c) use of scientific knowledge [2)

(ii) Potentiometer is based on null point method.Voltmeter draws current at the time ofmeasurement of potential difference across thecell. So, it dees not give the correct value of emf.For more precise measurement, it should below. [2]

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