grouping of capacitors

5/4/2014 Grouping of capacitors:

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Grouping of capacitors:

�� (i) If many capacitors are end to end, so that charge on each capacitor will be same then capacitors are said to be connected in Series.

�� In series combination, sum of potential of each capacitor be equal to the potential of battery.

��

��

V = V1 + V2 + V3 + ��.+Vn

��

��

��

��

�� Hence, effective capacitance in series combination is less then the least of the individual capacitances.

�� If tow capacitor connected in series combination then �

��

��

��

�� If c1 = c2 then

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(ii) If many capacitors are connected between the same two terminals, so that the potential difference across everyone of them is same, then they are said to be connected inparallel.

��

�� The sum of charge on the capacitors be equal to the charge supplied by battery.

�� q= q1 + q2 +q3

�� cv = c1v+c2v+c3v

��

or c= c1 + c2 + c3

(iii) If cp is the effective capacity when n identical capacitors are connected in parallel and Cs is their effective capacity when connected in series, then

��



(iv) If four identical plates are connected between point A and B as shown then it be equal to three capacitor connected in parallel combination. As shown

Hence equivalent capacitance between A and B �

�� Ce = 3C0

��

��

�� (v) If four identical plates are connected as, shown, then it be equal to two capacitor connected in parallel. Hence equivalent capacitance �

��

��

�� Ce= 2C0

��

�� (vi) If four identical plates are connected as shown then equivalent capacitance �

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Questions and Answer

Q1.�� Find equivalent capacitance between point A and B in each following problem.



Q2.�� Find in the given circuit �

��

�� (i) charge on each capacitor

�� (ii) Potential drop along each capacitor�� (iii) Potential difference between point A and B.

Ans.�� (i) The given capacitors are connected in series. Hence charge on each capacitor will be same. If equivalent capacitance C then -

��

��

�� C = 2 m F

�� If charge be q then �

�� q = cv = 2 x 12 = 24 mc�� Hence charge on each capacitor be 2c.

(ii) Since, capacitance of each capacitor be same and charge on each capacitor also be same. Hence, potential drop along each capacitor also be same.

��

�� V = 4 volt

(iii) Note -�� To find the potential difference between any two point, we start to move from one point to another point and write all the potential

difference with sign. if we go from positive plate to negative plate of a capacitor or battery then potential difference along capacitor or battery represented by negative (-)sign and vice versa. If we go through a resister in the direction of current then potential drop along the resistor is represented by negative (-) sign and vice versa.

�� Let potential of point A and B be respectively VA and VB. We short to move from point A to B clockwise.

��

��

�� or�� VA � VB = 8 volt

��

Second Method: -

�� If we move from point A to B anti clockwise then �

�� VA -12 + 24/6 = VB

�� VA � VB = 8 volt.

Q.3�� In the given circuit, find charge on each capacitor and potential difference, between point A and B.

Ans.�� (i) In the given figure both capacitors be in series because charge does not divide Hence charge on both capacitor will be same. Since polarity of batteries be

opposite. Hence net potential difference in the circuit �



��

�� V = 12 � 6 = 6 volt

�� The given circuit also drawn as �

��

�� If charge on capacitors be q then �

�� q= cv

�� = 8/3 x 6 = 16

�� q = 16 m c

(ii) Let potential of point A and B be respectively VA and VB.

To find potential difference between point A and B, we start to move from A clockwise and to the point B.

��

�� VA � VB = 10 volt

Q5.�� In the given figure, point Q connected with earth. Find potential or point P and R.

��

Ans.�� Since, point Q connected with earth, so potential of point Q be zero. All capacitor be in series and equivalent capacitance be (c/4). If charge supplied by battery be q

then -��

�� q = c/4 x 10�� { q = cv }

�� q = 2.5c

�� If potential of point�� P be VP then

��

��

��

�� VP � 5 = VR



�� or�� VR = + 5 volt

Q6.�� Find charge on the given capacitor.

��

Ans.�� The potential difference along the capacitor be equal to the potential difference along Resistance 50 W, and the branch in which capacitor exist, the current does notflow in that branch.

�� Hence, a down fig drawn as -

��

�� If current be I then �

��

��

�� P. D. along he resistor 50 W �� = iR

�� = 0.3 x 50

�� = 15 volt

�� Hence, charge on the capacitor = cv�� = 2 x 15

�� = 30 mc

Q7.�� Find the charges on the three capacitors shown in figure

��

Ans.�� Let the charges in three capacitors be as shown in figure�� Charge supplied by 10v battery is q1 and that from 20v battery is q2 then �

�� q1 + q2 + q3 ��.. (1)

This relation can also be obtained in a different manner. The charges on the three plates which are in contact add to zero. Because these plates taken together from an isolatedsystem which can�t receive charges from the batteries Thus.

�� q3 = q1 � q2 = 0

�� or�� q3 = q1 + q2

�� Applying second law in loops BCFAB and CDEFC we have

��

�� or�� q3 + 3q1 = 60

��

�� or�� 3q2 + 2q3 = 240

�� solving the above three eq. we have

��



��

��

�� q3 = 50 mC

�� and

�� Thus, charges on different capacitors are as shown in fig

Q8.�� What charges will flow through A, B and C in the direction shown in figure when switch s is closed

��

Ans.�� Let us draw two figures and find the charge on both the capacitors before closing the switch and after closing the switch.

��

��

Refer figure (a) when switch is open: Both capacitors are in series. Hence their equivalent capacitance is

��

�� There fore, charge on both capacitors will be same. Hence, using q = cv, we get

�� q= (30 + 60) (6/5)mc = 108mc

Refer figure (b) when switch is closed: Le q1 and q2 be the charges (in mc) on two capacitors. Then applying second law in upper and lower loop we have.

��

��

charge q1 and q2 can be calculated alternatively by seeing that upper plate of 2mF capacitor is connected with positive terminal of 30v battery. Therefore they are at the same

potential. Similarly, the lower plate of this capacitor is at the same potential as that of the negative terminal of 30v battery. So we can say that P.D. across 2F capacitor is also

30v

�� q1 = (c) (P.D.) = (2) (30) mc = 30 mc

�� Similarly, P.D. across 3 mF capacitor is same as that between 60 v batteries hence

�� q1 = (3) (60) mc = 180 mc

Now let qA charge flow from A in the direction shown. This charge goes to the upper plate of 2mF capacitor. Initially it has a charge +q and finally charge on it is + q1

Hence

q1 = q + qA

qA = q1 � q = 60 � 18 = -48 mc

similarly, charge qB goes to the upper plate of 3 mF capacitor and lower plate of 2mF capacitor initially both the plates had a charge +q,� - q or zero. And finally they have a

charge (q2 � q1) Hence

(q2 � q1) = qB + 0

qB = q2 � q1 = 180 � 60

120 mc

charge qc goes to the lower plate of 3mF capacitor. Initially it has a charge �q and finally � q2. Hence

�� -q2 = (-q) + qC

\�� qc = q � q2 = 180 � 180

�� = -72 m c



�� so, the charges will flow as shown below

��

Heat Generation

��

By heat generation we mean that when the switch is shifted (as discussed above) from one position to the other what amount of heat will be generated in the circuit. Such

problems can be solved by simple energy conservation principle.For this remember that when a charge �q flows from negative terminal to the positive terminal inside a battery of emf v is supplies a energy

�� E = qv]

�� and if opposite is the case i.e. charge +q flows in opposite direction then it consumes energy by the same amount

Now from energy conservation principle we and find the heat generated in the circuit in shifting the switch. Therefore

Heat generated = energy supplied by the battery

�� - energy consumed by the

battery + �Ui - �Uf

Here �Ui �� = energy stored in all the capacitors initially and

�Uf�� = energy stored in all the capacitors finally

For instance in the above example energy is supplied by 60v battery and consumed by 30v battery using E = qv, we have

�� Energy supplied�� = (72 x 10 -6) (60)

�� = 4.32 x 10-3J

�� Energy consumed �� = (48 x 10-6) (30)

�� = 1.44 x 10-3 J

��

��

�� = 6.3 x 10-3 J

�� \ Heat generated = (4.32 � 1.44 + 4.86 � 6.3) x 10-3J

�� = 1.44 x 10-3J

Q9.�� Find equivalent capacitance between points A and B shown in figure

��

Ans.�� Process � Give arbitrary potentials (v1, v2 �..etc.) to all terminals of capacitors/resistors. But notice that the point connected directly by a connected directly by a

connected directly by a conducting wire will have at the same potential. The capacitors/resistors having the same P.D. are in parallel. Make a table corresponding to the figure.

Now corresponding to this table a simplified figure can be formed and from this figure ceq and Req can be calculated.

Three capacitors have P.D. V1 � V2. so they are in parallel. Their equivalent capacitance is 3c.



Two capacitors have P.D. V2 � V3 so their equivalent capacitance is 2c and lastly there is one capacitor across which P.D. is V2 � V4. So let us make a table corresponding

this information

�� P.D.�� Capacitance�� V1 � V2�� 3C

�� V2 � V3�� 2C

�� V2 � V4�� C

Now corresponding to this table. We make a simple figure as shown below

As we had to find the equivalent capacitance between points A and B, across which P.D. is V1 � V4 from the simplified figure we can see that the capacitor of capacitance 2c

is out of the circuit and points A and B are as shown. Now 3c and c are in series and their equivalent capacitance is

Q10.�� In the circuit shown in the figure find the steady state charges on both the capacitors.

��

Ans.�� How to proceed � In steady state a capacitor offer an infinite resistance. Therefore the two circuits ABGHA and CDEFC have no relation with each other. Hence the

battery of emf 10 v is not going to contribute any current in the lower circuit. Similarly, the battery of emf 20 v will not contribute to the current in the upper circuit. So first we will

calculate the current in the two circuits, then find the potential difference VBG and VCF across the capacitors to find the charge stored in them

�� Current in the upper circuit i1

��

�� VBG = VB � VG = 3 i1�� = 3 x 2 = 6 volt

��

�� current in the lower circuit i2 =

��

�� VCF = VC � VF = 4i2 =- 4 x 2 = 8 volt

�� charge on both the capacitors will be same Let it beq. Applying Kirchhoff�s second law in loop



��

��

�� q = 4 x 10-6 c

�� q = 4 mc �� Ans.

Q11.�� Five identical capacitor plates each of area A are arranged such that adjacent plates are at distanced apart. The plates are connected to a source of emf v as shown isfig what is the magnitude and nature of charge an plate 1 and 4 respectively �

Ans.�� These five plates constitute four identical capacitors in parallel each of capacity

�� Now as plate � 1 is connected to positive terminal of battery and is a part of one capacitor only so charge on it

��

��

�� How ever, the plate -4 is connected to negative terminal of the battery and is common to two identical capacitors (unparallel)

��

�� [as qc = qD = -q1]

Q12.�� A capacitor is filled with two dielectrics of the same dimensions but of dielectric constant 2 and 3 respectively. Find the ratio of capacities in the two possible

arrangements.

Ans.�� If a capacitor is filled with two dielectrics with same dimension, only following two arrangements are possible

��

�� Arrangement (A) is equivalent to a combination of two capacitors each of area A and separation (d/2) in series i.e.

��

��

�� And the arrangement (B) is equivalent to a combination of two capacitors each of area (A/2) and separation d in parallel i.e.

��



��

��

��

�� so that

��

�� Here K1 = 2 and K2 = 3

��

Q13.�� The capacitance of a variable radio capacitor can be change from 50pF to 950 pF by turning the dial from 00to 1800 with the dial set at 1800 the capacitor is connect to a

400 v battery After charging, the capacitor is disconnected from the battery and the dial is turned to 00 (a) what is the potential difference across the capacitor when the deal real

00? (b) How much work is required to turn the dial if friction is neglected?

Ans.�� When the dial is at 1800 the capacity is 950pF so the charge stored by the capacitor

�� q = cv = (950 x 10-12) x (400) = 3.8 x 10-7 c

(a) Now when battery is disconnected, capacitor becomes isolated and so charge on it will remain unchanged, since in turning the dial from 1800 to 00 capacity changes form

950 pF to 50 pF from �

�� q = q1 � i.e. �� cv = c� v�

��

�� (b) work done in changing the position of dial from 1800 to 00

��

��

��

Q14.�� Two parallel plate capacitors of capacity c and 2c are connected in parallel and charged to a potential difference v. The battery is then dis connected and the space

between the palates of the capacitor c is completely filled with a material of dielectric constant K. What is the final potential difference across the capacitor?

Ans.�� As q = cv, the charge on capacitors, when they are charged in pavallel by a source of potential difference v

��

�� q = q1 +q2= cv +(2c)v = 3cv

�� Now with the removal of battery system becomes isolated i.e.�� q� = q � i.e. q�1 + q�2 = q1 + q2 = 3cv

�� And as with the introduction of dielectric capacity becomes k times

�� c' = c�1 + c�2 = kc + 2c = (k + 2)c

��



�� So the find potential

��

Q15.�� Fig. Shows two identical parallel plate capacitor connected to a battery with the switch s closed. The switch is now opened and the free space between the plates of thecapacitors is filled with a dielectric of K = 3. Find the ratio of the total electrostatic energy stored in both the capacitors before and after the introduction of the slab

Ans.�� When the switched s is the two capacitors in parallel will by the same potential difference V, so

��

�� When switch S is opened and dielectric is introduced fig (B)

��

�� (i) capacity of both the capacitor becomes K times� = i.e. c�1 = c�2 = kc�� i.e.

�� (ii) capacitor A remains connected to battery

�� v�1 = v�� i.e.

�� (iii) capacitor B becomes isolated i.e.

�� q�2 = q2 i.e.�� c�2 v�2 = c2 v2�� [as q = cv]

�� or (kc) v�2 = cv�� i.e. v�2 = (v/k) [asc�2 = kc]

�� so final energy

��

�� UF =

�� So from equation (1) and (2)

��

Q16.�� In the diagram shown in fig. Find the potential difference between the points A and B between points B and C in steady state

Ans. �� In this problem

(i) As there is no closed path, in steady state the current in every mesh in the circuit is zero and hence in steady state resistance are in effective. (The function of theseresistances is to limit the current in the circuit during charging and protect the capacitors and battery from getting damaged)

�� (ii) Point A, P, Q, and R are at same potential while C, N, M and L are at another same potential.

�� In the light of above facts simplified equivalent circuit becomes as shown in fig

��



��

In this circuit in the arm PL (3mF + 3mF) is in series with (1mF + 1mF) i.e. 6mF and 2mF are in series with 100v PD across AC. Since in series potential divides white charge

remains same�� V= V1 + V2 �� and q1 = q2 i.e. C1V1 = C2V2

�� i.e. VAB + VBC = 100v and 6VAB = 2VBC

�� solving these for V AB and V BC we find that

�� VAB = VA � VB = 25 v and� VBC = VB � VC = 75v

EXERCISE - 2 Q1.�� Three capacitors of capacitances 12 mF each are available. The minimum and maximum capacitances which may� e obtained form there are

�� (i) 12 mF, 36 m F�� (ii) 4 m F, 12 m F�� (iii) 4 m F, 36 m F

�� (iv) 0 m F, � m F

Ans.�� (iii)

Q2.�� In the given figure, there capacitor c1, c2 and c3 are joined to a battery, with symbols having their usual meanings, the correct conditions will be

��

�� (i) Q1 = Q2 = Q3 AND V1 = V2 = V3 + V

�� (ii) Q1 = Q2 + Q3 and V= V1 + V2 + V3

�� (iii) Q1 = Q2 +Q3 and V = V1 + V2

�� (iv) Q3 = Q2 and V2 = V3

Ans.�� (iii)

Q3.�� An infinite number of identical capacitors each of capacitance 1 mF are connected as shown in the figure. Then the equivalent capacitance between A and B is

��

�� (i) 1 m F

�� (ii) 2mF�� (iii) =1/2 mF�� (iv) �



Ans.�� (ii)

Q4.�� The identical parallel plate capacitor are placed in series and connected to a constant voltage source of V0 volt. If one of the capacitors is completely immersed n

a liquid with dielectric constant K The potential difference between the plates of the other capacitor will change to

��

��

��

��

Ans.�� (ii)

Q5.�� Six equal capacitors each of capacitance c are connected as shown in the figure. Then the equivalent capacitance between A and B is

��

�� (i) 6c�� (ii) c�� (iii) 2c

�� (iv) c/2

Ans.�� (iii)

Q6.�� IN given circuit diagram E = 5 volt r = 1W, R2� = 4W, R1 = R3 = 1W and c = 3mF. then the numerical value of the change on each plate of the capacitor is

��

�� (i) 24 mc

�� (ii) 12 mc�� (iii) 6 mc

�� (iv) 3 mc

Ans.�� (iii) Q7.�� A battery of e.m.f. v volts, resistors R1 and R2 a condenser c and switch S1 and S2 are connected in the circuit as shown in the figure below

��

�� The condenser will get fully charged to v volt then�� (i) S1 and S2 ae both closed (ii) S1 and S2 are both open (iii) S1 is open and S2 is closed (iv) S1 is closed and S2 is open

Ans.�� (iv)

Q8.�� In the circuit shown in the figure, each capacitor has a capacity of 3 mF. The equivalent capacity between A and B is



��

�� (i) � mf�� (ii) 6 mF

�� (iii) 3 mF�� (iv) 5 mF

Ans.�� (iv)

Q9.�� A circuit has a section AB as shown in the figure If the potential difference between points A and B is v volts, then potential difference across c1 is

�� (i) (V � E) C1

�� (ii) (V + E) C2

��

��

Ans.�� (iii)

Q10.�� For the circuit shown in the adjoining figure, the charge on 4 mF capacitor is

��

�� (i) 30mc�� (ii) 40mc

�� (iii) 24 mc�� (iv) 54 mc

Ans.�� (iii)

Q11.�� A number of capacitors each of capacitance 1 mF and each one of which get punctured if a potential difference just exceeding 500 volt is applied, are provided. Thenan arrangement suitable for giving a capacitor of 2 mF across which 3000 volt may be applied requires at least

�� (i) 18 component capacitors�� (ii) 36 component capacitors

�� (iii) 72 component capacitors�� (iv) 144 component capacitors

Ans. (iii)

Q12.�� A capacitor of capacitance 1 m F with stands a maximum voltage of 6 kv. While another capacitor of capacitance 2 mF the maximum voltage 4 kv. If they are connectedin series, the combination can with stand a maximum of.

�� (i) 6 Kv�� (ii) 4 kv�� (iii) 10 kv

�� (iv) 9 kv

Ans.�� (iv)



Q13.�� Seven capacitors each of capacitance 2 mF are to be connected to obtain a capacitance of (10/11 mF) which of the following combination is possible

�� (i) 5 in parallel, 2 in series�� (ii) 4 in parallel, 3 in series

�� (iii) 3 in parallel, 4 in series�� (iv) 2 in parallel, 5 in series

Ans.�� (i)

Q.14.�� Two capacitor 2 mF and 4 mF are connected in parallel. A third capacitor of 6 mF capacity is connected in series. The combination is then connected across a 12 vbattery. The voltage across 2 mF capacity is

Ans.�� (ii)

Q15.�� Seven capacitors, a switch s and a source of e.m.f. are connected as shown in the figure. Initially, s is open and all capacitors are uncharged. After s is closed and

steady state is attained the potential difference in volt across the plates of the capacitor A is

��

�� (i) 12�� (ii) 15

�� (iii) 17�� (iv) 19

Ans.�� (i)

Q16.�� Four identical capacitors are connected in series with a 10v battery as shown. The point N is earthed the potentials of point A and B are

�� (i) 10v, 0v

�� (ii) 7.5v, - 2.5v�� (iii) 5v, -5v

�� (iv) 7.56v, 2.5v

Ans.�� (ii) Q17.�� Your are given 32 capacitors each having capacity 4 mF. How do you connect all of them to prepare a composite capacity having a capacitance 8 mF

�� (i) 2 condensers in series and 16 such groups in parallel�� (ii) 8 condensers in series and 4 such groups in parallel

�� (iii) 4 condensers in series and 8 such groups in parallel�� (iv) all of them in series.

Ans.�� (iii)

Q18.�� A parallel plate capacitor of Area A. plate separation d and capacitance c is filled with three different dielectric materials having dielectric constant K1, K2 and K3 as

shown. If a single dielectric material is to be used to have the same capacitance c in this capacitor. Then its dielectric constant K is given by (A = Area of plates)

��



��

��

��

��

Ans.�� (ii)

Q 19.�� The numerical value of the charge of n either plate of the capacitor c shown in the figure is

��

�� (i) CE

��

��

��

Ans.�� (iii)�� Q20.�� Three capacitors are connected to a Dc source of 100 volts as shown in figure. If the charges accumulated on the plates of C1, C2 and C3 are qb, qd, qe and qf

respectively then

�� (i) qb + qd +qf = (100/9) coulomb

�� (ii) qb + qd + qf = o coulomb�� (iii) qa + qc� +qc = 50 coulomb (iv) pb = qd = qf

Ans.�� (iv)

Q 21.�� Six metallic plates each with a surface area of one side A, are placed at a distance from each other. The alternate plates are connected to paints P and Q as shown in

figure. The capacitance of the system is

��

��

��



��

��

Ans. (ii)

Q22.�� Four capacitors are connected in a circuit as shown in figure. The effective capacitance between points A and B will be

��

�� (i) 28/9 mF

�� (ii) 5 mF�� (iii) 4 mF

(iv) 18 mF

Ans.�� (ii)

Q23.�� Three plates A, B, C each of area 50 cm2 have separation 2 mm between A and B and 3mm between B and C. The energy stored when the plates are fully charged is

��

�� (i) 2 m J�� (ii) 1.6 m J

�� (iii) 5 m J�� (iv) 7 m J

Ans.�� (ii)

Q24.�� Four capacitors and a switch share connected to a source as shown in the figure. Initially s is open and the capacitors are uncharged after s is closed and steady state

is reached. The energy stored in 4 m F capacitor in the units of 10-5 J is

��

�� (i) 40�� (ii) 30

�� (iii) 20�� (iv) 15

Ans.�� (iii)

Q25.�� Two batteries, two resistors and two condensers are connected as shown in the figure. The charge on 2m F capacitor is

��



�� (i) 30 mC�� (ii) 20 mC

�� (iii) 25 mC�� (iv) 48 mC

Ans.�� (i)

Q 26.�� In the circuit diagram, potential difference between points A and B is 200 volt the potential difference between points A and B when the switch s is open, is

��

�� (i) 100 volt

�� (ii) 200/3 volt

�� (iii) 100/3 volt

�� (iv) 50 volt Ans.�� (ii)

Q27.�� Two capacitor of capacitance C1 = 2 m F and C2 = 3 m F are connected as shown to two batteries A and B which have e.m.f 6 v and 10 V respectively. Energies stored

in the capacitors C1 and C2 are

�� (i) 36 mJ, 36 mJ�� (ii) 24 mJ, 24 mJ�� (iii) 24 mJ, 36 mJ

�� (iv) 36 mJ, 24 mJ

Ans.�� (iv)

Q28.�� In the Q.27, what is the stored energy in the two capacitors C1 and C2 if the terminals of battery A are reversed

�� (i) 36 mJ . 54 mJ�� (ii) 54 mJ . 84 mJ

�� (iii) 24 mJ .� 36 mJ�� (iv) 36 mJ . 24mJ

Ans.�� (iv)

Q29.�� In the Q.27 capacitors C1 and C2 if the battery B is disconnected and the points X and Y are connected together

�� (i) 36 m J, 24 mJ�� (ii) 24 m J, 54 mJ

�� (iii) 36 mJ, 54 mJ�� (iv) 72 mJ, 108 mJ

Ans.�� (iii)

Q30.�� Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch S. If now the switch S is opened and the

space between the plates filled with a dielectric of specific inductive capacity of relative permittivity 3 then

�� (i) The potential difference as well as charge on each capacitor goes down by a factor of 3

�� (ii) The potential difference as well as a charge on each capacitors goes up by a factor of 3.�� (iii) The potential difference across B remains constant and charge on A remains uncharged

�� (iv) The potential difference across A remains constant and charge on B remains uncharged

Ans.�� (iv) Q. 31.�� Three uncharged capacitors of capacitance C1, C2 and C3 are connected as shown in figure to one another and to points A, B and D at potential VA, VB, And VD. Then the

potential at 0 will be



��

��

��

��

��

Ans.�� (ii)��

Q32.�� Five identical capacitor plates each of area A, are arranged such that the adjacent plates are at a distance d apart. The plates are connected to a battery of e.m.f. E voltas shown. The charge on plates 1 and 4 respectively are.

��

��

��

��

��

Ans.�� (i)

Q33.�� The capacities of three capacitors are in the ratio 1: 2: 3 their equivalent capacity when connected in parallel is F more than that when they are connected in series. The

individual capacitors are of capacities in mF�� (i) 4, 6, 7�� (ii) 1, 2, 3

�� (iii) 2, 3, 4�� (iv) 1, 3, 6

Ans.�� (ii)

Q34.�� In the circuit shown in the figure c =6 mF The charge stored in the capacitor of capacity is

��

�� (i) zero�� (ii) 90 mC

�� (iii) 40 mC�� (iv) 60mC



Ans.�� (iii)

Q35.�� Two identical capacitors 1 and 2 are connected in series to a battery as shown in the figure capacitor 2 contains a dielectric slab of dielectric constant K as shown. Q1

and Q2 are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are Q1 and Q2 then �

��

��

��

��

��

Ans.�� (iii) Q36.�� In the circuits as shown in the figure

��

�� The effective capacitance between A and B is�� (i) 3 m F

�� (ii) 2 m F�� (iii) 4 m F

�� (iv) 8 m F

Ans.�� (iii) Q37.�� What is the equivalent between point A and B.

��

�� (i) � C

�� (ii) 4/3 C�� (iii) 3C

�� (iv) 4C Ans.�� (i)

Q38.�� What in the equivalent capacitance between point A and B.

��

�� (i) 3/5 C�� (ii) 5/3 C�� (iii) 3C



�� (iv) 5C Ans.�� (iii)

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