chapter10pp090-101

8/10/2019 Chapter10pp090-101

1/12

Worksheet

Worked examplesPractical 1: The I/Vcharacteristics of components

Practical 2: The effect of temperature on the resistanceof a thermistor

Practical 3: Identifying a material from its resistivityEnd-of-chapter test

Marking scheme: Worksheet

Marking scheme: End-of-chapter test

Chapter 10

Resistance and resistivity

OCR (A) specifications: 5.2.1h,i,j,k,l,m,q

8/10/2019 Chapter10pp090-101

2/12

A

6.0 V

S

I

0

0

I(mA)

Time

60

20

10 Resistance and resistivity Cambridge University Press 2005 91

WorksheetIntermediate level1 Define electrical resistance. [1]

2 State Ohms law. [1]3 Write a word equation for the resistance of a length of metal wire in terms of

the resistivity of the metal, the length of the wire and its cross-sectional area. [1]

4 A component is connected to a d.c. supply. The supply has negligible internalresistance. At 6.0V, the current in the component is 0.023A. When the p.d. is

doubled, the current in the component increases to 0.100A.

a Calculate the resistance of the component at 6.0V. [2]

b Does the component obey Ohms law? Explain your answer. [2]

5 The diagram below shows theI/Vcharacteristics of two componentsAand B.

The components are connected in series to a battery. The current in each

component is the same and equal to 0.60A.

Calculate the individual resistances ofAand B. [2]

6 A 14m long copper wire of cross-sectional area 4.2 108m2 is wound into a coilfor a loudspeaker. The resistivity of copper is 1.7 108m. Calculate the resistance

of the wire. [3]

Higher level7 The diagram shows a thermistor connected to a d.c. supply.

The supply has negligible internal resistance. When the switch S is closed, the

currentI in the circuit changes as shown in the graph on the right.

0

0

0.20

V (V)2 4 6

0.40

0.60

0.80

1.00

1 3 5

I(A)

A B

8/10/2019 Chapter10pp090-101

3/12

92 Cambridge University Press 2005 10 Resistance and resistivity

a Explain why the current changes in the manner shown in the graph. [2]

b Calculate the ratio:

minimum resistance of thermistor

maximum resistance of thermistor[2]

8 The resistance across the ends of a 15cm long pencil lead is 3.6 . Calculate itsradius given that the pencil lead material has a resistivity of 7.5 105 m. [3]

9 A piece of metal is shaped into a rectangular block as shown below.

The metal has a resistivity of 4.3 104 m.

a The resistance of the block depends on which pair of faces it is measuredbetween. Calculate the minimum resistance between two opposite faces of

the block. [4]

b What is the maximum current in the block in awhen connected to a 0.050V

supply of negligible internal resistance? [2]

10 A filament lamp is connected to a d.c. supply. The current in the lamp is 2.0A whenthe potential difference across it is 12V. When operating at 12V, the filament of the

lamp has a cross-sectional area of 4.9 109 m2 and the resistivity of the filament

material is 5.6 107 m. Calculate the length of the filament in centimetres. [4]

Extension11 A glass tube contains a conducting liquid of length 5.0cm. The internal radius of

the tube is 1.4cm. The resistivity of the liquid is 8.5 105 m. The liquid is poured

onto a horizontal surface and quickly sets in the form of a uniform cylindrical disc

of radius 25cm. Calculate the resistance of this disc across its two opposite larger

surfaces. You may assume that the resistivity of the material remains constant. [4]

12 The resistivity of aluminium is twice that of copper. However, the density ofaluminium is one-third that of copper.

a For equal length and resistance, calculate the ratio:

mass of aluminium

mass of copper [3]

b Use the Internet to investigate the construction of power cables used for the

National Grid. You may be surprised to find that the current-carrying cables are

made from aluminium and not copper. Explain why this is so.

0.60 cm

0.80 cm

4.0 cm

Total: 36

Score: %

8/10/2019 Chapter10pp090-101

4/12


Worked examplesExample 1A semiconductor diode is connected to a variable d.c. supply of negligible internal

resistance. The current in the diode is zero when the p.d. across it is 0.40 V. The current

increases to 30mA when the p.d. across the diode is 0.65V. Calculate the resistance ofthe diode at 0.40V and 0.65V. Does the diode obey Ohms law?

The resistanceR of a component is given by:

R =V

I

At 0.40V, the currentIis 0A. Therefore:

R =0.40

= 0

(The diode is not conducting; hence its resistance is infinite.)

At 0.65V, the currentIis 30mA. Therefore:

R =0.65

= 22 0.03

(The resistance of the conducting diode is quite small.)

The resistance of the diode changeswith p.d. (or current). Therefore, the current cannot

be directly proportional to the p.d. The diode does not obey Ohms law.

Remember that any number divided by zero is infinite.

Remember to convert the current into amperes.

Failure to do this will give the resistance in k.

8/10/2019 Chapter10pp090-101

5/12


Tip

You do not need to use algebra to get the answer. You can first calculate the

resistivity of the wire and then use this to calculate the resistance of the wire.

The resistivity is given by:

=RA

l

R = 14 l = 3.2m A = r2 = (1.5 104)2 = 7.07 108 m2

=14 7.07 108

= 3.09 107 m3.2

For the new wire:

l = 3.2 5 = 16m A = r2 = (2 1.5 104)2 = 2.83 107m2 = 3.09 107m

Its resistanceR is therefore:

R =l

=3.09 107 16

= 17.5 18A 2.83 107

Example 2A wire of radius 0.15mm and length 3.2m has a resistance of 14. Calculate the

resistance of a wire made from the same material but having five times the length and

twice the radius.

The resistanceR of the wire is given by:

R =l

A

The cross-sectional areaA of the wire is given by:

A = r2

where ris the radius of the wire.

Therefore:

R = l

l

r

2

r

2

The resistance of the wire will therefore increase by a factor5

= 1.25. The resistance of22

the new wire is:

R = 1.25 14 = 17.5 18

8/10/2019 Chapter10pp090-101

6/12


Practical 1The I/V characteristics of componentsSafety

Always take sensible safety precautions when using mains-operated supplies. Teachersand technicians should follow their school and departmental safety policies and should

ensure that the employers risk assessment has been carried out before undertaking any

practical work.

Apparatus

variable d.c. supply digital ammeter

1m of 40swg nichrome wire digital voltmeter

filament lamp (60mA, 6 V) 100 resistor (for diode experiment)

silicon diode connecting leads

Introduction

You can identify a component from itsI/Vcharacteristics. In this experiment you will

determine theI/V characteristics of a metallic wire kept at a constant temperature, a

filament lamp and a semiconductor diode. You will find information on the

components mentioned above on pages 90 and 91 of Physics 1.

Procedure

The diagrams show appropriate circuits for investigating the different components. For

the diode experiment, it is vital to include a safety resistor.

1 Set up the appropriate circuit for the component you are investigating.

2 Change the potential difference across the component from zero to 6.0V in steps of0.5V.

3 Measure the current for each p.d.

4 Record your results in a table.

5 On the same axes, plot a current against voltage graph for each of the components.(You should be able to identify the component from the specific shape of the

I/V graph.)

Guidance for teachers

One metre of the nichrome wire has a resistance of about 90 . Using a 1m length makes

it possible to plot theI/Vgraphs for all the components on the same axes for comparison.

If a variable d.c. supply is not available, then a rheostat may be used as a potential divider.

A

V

A

V

VITAL SAFETY RESISTOR

100 diode

filament lamp or metallic wire

8/10/2019 Chapter10pp090-101

7/12


Practical 2The effect of temperature on the resistance of a thermistorSafety

Take care when pouring the boiling water into the beaker. Teachers and techniciansshould follow their school and departmental safety policies and should ensure that the

employers risk assessment has been carried out before undertaking any practical work.

Apparatus

6.0V battery thermometer

100ml beaker digital voltmeter

NTC thermistor digital ammeter

plastic bag connecting leads

electric kettle

Introduction

In this experiment you will investigate the effect that temperature has on the resistanceof a negative temperature coefficient (NTC) thermistor. Their resistive properties are

described on page 93 ofPhysics 1.

Procedure

The circuit shown here may be used to investigate the

behaviour of a thermistor.

1 Put the thermistor in a waterproof plastic bagand place it into a beaker.

2 Pour boiling hot water into the beaker.

3 Stir the water and measure the temperature ofthe water, the potential difference Vand the

currentI.


5 For every 10 C drop in temperature, measureIand V.

6 Calculate the resistanceR of the thermistorat each temperature using the equation:

R =V

I

7 Repeat the experiment twice and determine the average resistance at eachtemperature.

8 Plot a graph of resistanceR against temperature .

Guidance for teachers

The specifications only require knowledge of NTC thermistors. However, if there is time,

selected students could also determine the properties of PTC thermistors.

A

V

6.0 V

beaker withhot water

thermometer

8/10/2019 Chapter10pp090-101

8/12


Practical 3Identifying a material from its resistivitySafety

Do not attempt to measure the current for zero length. This will short out the batteryand send a potentially damaging current through the ammeter. Teachers and

technicians should follow their school and departmental safety policies and should

ensure that the employers risk assessment has been carried out before undertaking any

practical work.

Apparatus

6.0V battery (or d.c. supply) crocodile clip

manganin or eureka or nichrome wire digital voltmeter

ruler digital ammeter

micrometer connecting leads

IntroductionIn this experiment you will determine the resistivity of a metal and identify it by using

either a databook or the Internet.

Procedure

The diagram shows an arrangement that may be used

to determine the resistivity of a metal. You may use any

available wires in the laboratory.

1 Measure the diameter of the wire at differentpoints. Use the average diameter d to determine

the cross-sectional areaA of the wire using:

A =d2

4

2 For a 10cm long wire, measure the currentIandthe potential difference V.


4 Calculate the resistanceR of the wire using:

R =V

I

5 Increase the length of the wire in steps of 10cm and determine

the resistance for each length.6 Plot a graph of resistanceR of the wire against its length l.

7 Draw a straight line of best fit through the data points.

8 Find the gradient of the line, which is equal to

,A

where is the resistivity of the metal.

9 Calculate the resistivity using the relationship:

= gradient A

10 Use the Internet or a science databook to identify the metal.

A

V

6.0 V

crocodile clip

l

resistance wire

0

0

R

l

gradient =

A

8/10/2019 Chapter10pp090-101

9/12


End-of-chapter testAnswer all questions.

1 Name a non-ohmic component. [1]

2 The graph shows theI/V

characteristics of two types offilament lampsX andY.

a Does lampX obey Ohms law?

Explain your answer. [1]

b Calculate the resistance of

lampX at 2.0V. [3]

c Describe how the resistance

of lampYdepends on the

current. [1]

3 A negative temperature coefficient (NTC) thermistor and its connecting leads arecoated with a high-resistivity plastic material. The thermistor is placed in a beaker

containing hot water. The temperature of the water is kept constant at 80 C. TheI/V

characteristic of the thermistor is shown below.

a Calculate the resistance of the thermistor. [2]

b State and explain the change, if any, to the shape of theI/Vgraph when the

temperature is lowered and maintained at 30 C. [2]

4 A wire is made of a material of resistivity . Write an equation for theresistanceR of a wire of length l and diameter d. [2]

5 According to a databook, a manganin wire of radius 0.15mm has a resistance of5.33 per metre of length.

a Calculate the resistivity of manganin. [4]

b Explain how your answer to awould change if the manganin wire had

twice the radius. [2]

0

0

20

V (V)2 4 6

40

60

80

100

1 3 5

I(mA)

YX

0

0

20

V (V)4 8 12

40

60

80

100

2 6 10

I(mA)

80 C

Total: 18

Score: %

8/10/2019 Chapter10pp090-101

10/12


Marking schemeWorksheet

1 Resistance =potential difference

current[1]

2 The current in a metallic conductor kept at a constant temperature is directlyproportional to the potential difference across its ends. [1]

3 Resistance =resistivity length

cross-sectional area[1]

V 6.04 a R =

I=

0.023[1]; R = 260 [1]

b At double the p.d., the resistance is

12R =

0.100= 120 [1]

The resistance is not constant. Therefore, the current cannot be directlyproportional to the voltage. The component is non-ohmic. [1]

1.0 3.55 RA=

0.60= 1.7 [1]; RB =

0.60= 5.8 [1]

6 R =l

[1]; R =1.7 108 14

[1]; R = 5.67 [1]A 4.2 108

7 a As current flows in the thermistor, its temperature increases. This causes adecrease in the resistance of the thermistor and therefore an increase in the

current (the p.d. is constant at 6.0V). [1]

Eventually, the temperature reaches a maximum and therefore the resistance of

the thermistor is lower but constant. The current is therefore also constant. [1]

6.0 6.0b Rmin =

0.060= 100 , Rmax =

0.020= 300 [1]

Ratio =Rmin

=100

=1

0.33 [1]Rmax 300 3

8 R =l

[1]; A =l

=7.5 105 0.15

= 3.13 106 m2 3.1 106 m2 [1]A R 3.6

A = r2 so r=3.13 106

= 9.97 104 m 1.0 103 m [1]

9 a Minimum resistance shortest length and largest cross-sectional area [1]

l = 0.60 cm, A = 4.0 0.80 = 3.2cm2 = 3.2 104 m2 [1]

R =l

=4.3 104 6.0 103

[1]; R = 8.1 103 [1]A 3.2 104

V 0.050b I=

R=

8.1 103[1]; I= 6.2A [1]

V 1210 R =

I=

2.0= 6.0 [1]

l =RA

[1]; l =6.0 4.9 109

= 0.053m [1]; l = 5.3cm [1] 5.6 107

11 Volume remains constant (r2

h = constant) [1]252 l = 1.42 5.0 (l = new length), so l = 1.57 102 cm = 1.57 104 m [1]

R =l

=8.5 105 1.57 104

[1]; R = 6.8 108 [1]A (25 102)2

8/10/2019 Chapter10pp090-101

11/12


12 a Mass = density volume (M = DV) [1]

Resistance remains the same, and equal length l, therefore:

Cu l=Al l

soAAl

=Al

= 2 [1]ACu AAl ACu Cu

Ratio =MAl

=DAlAAl l

=

(

DAl

) (

AAl

)=

1

2= 0.67 [1]

MCu DCuACu l DCu ACu 3 1

The mass (and hence weight) of aluminium overhead cables is 67% that of

copper for equal length and resistance.

b Internet research. Cables are typically aluminium with steel reinforcement.

8/10/2019 Chapter10pp090-101

12/12


Marking schemeEnd-of-chapter test1 Semiconductor diode or a filament lamp [1]

2 a The current is not directly proportional to the p.d., thereforeX does not obeyOhms law. [1]

b V= 2.0V andI= 60mA [1]

V 2.0R =

I[1]; R =

0.060= 33 [1]

c The resistance ofYdecreases as the current increases. [1]

V 123 a R =

I[1]; R =

0.080= 150 [1]

b The resistance of the thermistor is higher at a lower temperature. [1]

1SinceR =

gradient of line, the graph is a straight line of smaller gradient. [1]

4 R =l

and A =r2 =d2

[1]; R =l

=4l

[1]A 4 A d2

5 a l = 1.0m, r= 1.5 104 m,R = 5.33 [1]

=RA

=5.33 (1.5 104)2

[1]; = 3.8 107 [1]; Unit: m [1]l 1.0

b There is no change. [1] The resistivity depends on the material and not on its

dimensions. [1]

chapter10pp090-101

Documents