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Physics Support Materials
Higher Electricity and Electronics

Electric Fields and Resistors in Circuits
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20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
3

An electron volt is a unit of energy. It represents the change in
potential energy of an electron which moves through a potential
difference of 1 volt. If the charge on an electron is 1.6 x 10 -19 C,
what is the equivalent energy in joules?
E  QV
E  1.6 10 19 1
E  1.6 10 19 J
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
4

Mass of an electron = 9.1 x 10 -31 kg.
Charge on an electron = 1.6 x
10 -19 C. The electron shown below is accelerated across a p.d. of 500
V.
(a) How much electrical work is done?
W  QV
W  1.6 1019  500
W  8 1017 J
(b) How much kinetic energy has it gained?
Work done  Energy gained
EK  8 1017 J
1
17
(c) What is its final speed? 8  10
  9.110 31  v 2
2
1 2
17
EK  mv
2

8

10
14
2
v2 

1
.
76

10
9.110 31
v  1.33 107 m s -1
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
5

Electrons are ‘fired’ from an electron gun at a screen. The p.d.
across the gun is 2000 V. After leaving the positive plate the
electrons travel at a constant speed to the screen. Assuming the
apparatus is in a vacuum, at what speed will the electrons hit the
screen?
W  QV
W  1.6 10 19  2000
W  3.2 10 16 J
1 2
EK  mv
2
3.2 10
16
1
  9.110 31  v 2
2
16
2

3
.
2

10
14
v2 

7

10
9.110 31
v  2.65 107 m s -1 Click the mouse to continue
Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
6

What would be the increase in speed of an electron accelerated
from rest by a p.d. of 400 V?
W  QV
W  1.6 10 19  400
W  6.4 10 17 J
1
EK  mv 2
2
1
6.4 10 17   9.11031  v 2
2
17
2

6
.
4

10
14
v2 

1
.
41

10
9.110 31
v  1.2 107 m s -1
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
7

An X-ray tube is operated at 25 kV and draws a current of 3 mA.
(a) Calculate
(i) the kinetic energy of each electron as it hits the target
EK  QV EK  1.6 1019  25 103
EK  4 1015 J
(ii) the velocity of impact of the electron as it hits the target
1
EK  mv 2
2
4 10
15
1
  9.110 31  v 2
2
2  4 10 15
14
v 

88

10
9.110 31
v  9.4 107 m s -1
2
(iii) the number of electrons hitting the target each
3
second.
Q  It
Q  3 10 1
3 103
16
No. of electrons 

1
.
875

10
1.6 1019
(b) What happens to the kinetic energy of the
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electrons?
It is transferred to the Ek of the X ray photon
Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
11

In the circuit opposite:
(a)
what is the total
resistance of the
RP R1 R2
RP 12 4
12
4
circuit
P 
 
 

R
1
1 1
1
1 1
4
12

(b) what is the resistance between X
3
and Y

(c) find the readings on the ammeters
I
V
R
I
12
6
I  2 A (A1 )
 3
RS  3  3
RS  R1  R2
RS  6 
The current splits in the ratio of 12 : 4 or 3 : 1
The current through the 4  resistor is 3/4 of 2 A or 1.5 A

(d) calculate the p.d. between X and Y

(e) what power is supplied by the
battery?
P  12  2  24 W
P  VI
V  IR
V  23  6 V
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
12
The circuit opposite uses the 230 V alternating mains supply. Find
the current flowing in each resistor when:
(a) switch S is open
(b) switch
RP R1 R2 RP 8 24
24
S is closed.
S
S
1
2
 
 

R R R
R  20 
1
1 1 1 1 1
4

I
V
R
I
I  11.5 A
4
R 
 6
24
RS  R1  R2
RS  6  12  18 
230
20
P
V
230
I
 12.8 A 12  
18
R
Voltage across 12   IR  12.8 12  153.3 V
I
Voltage across parallel branch  230  153.3  76.7 V
V
76.7
V
76.7
(8 ) I 

 9.6 A
(24 ) I 

 3.2 A
R
8
R
24
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
13


An electric cooker has two settings, high and low. It takes 1 A at the
low setting and 3 A at the high setting.
(a) Find the resistance of R1
and R2
At the low setting, the switch is open
R1 
I
V

1
 230 
230
RP R1 R2
76.7 230 R2
At the high setting, the switch is closed
 


I
3
1
1
1
1
1
1
RP 

 76.7 
2
R2 76.7 230 R2 230
230
V
230
2 
R
 115 




230
1
1
1
1 3 1
2

(b) What is the power consumption at each
setting?
Open
P  VI
P  230 1  230 W
Closed P  VI
P  230  3  690 W
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
14

(a) Find the value of the series resistor which would allow the bulb
to operate at its normal rating.
Finding the current through the bulb
P  VI
I
P
V
I
36
 3A
12
The current through the bulb is the same as the current through the resistor
R

V
I
R
12
 4
3
(b) Calculate the power dissipated in the
resistor
P  VI
P  12  3  36 W
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
15

In the circuit below, r represents the internal resistance of the cell
and R represents the external resistance of the circuit. When S is
open, the voltmeter reads 2.0 V. When S is closed, it reads 1.6 V
and the ammeter reads 0.8 A.
2.0 V

(a) What is the e.m.f. of the cell?

(b) What is the terminal potential difference when S is
1.6 V
closed?
(c) Calculate the values of r and
V
1.6
R.
R
 2
R

I
E  IR  Ir

0.8
2  1.6  0.8  r
r
2  1.6
 0.5 
0.8
(d) If R was halved in value, calculate the new readings on the
ammeter and voltmeter.
E  I (R  r)
2  I (1  0.5)
2
I
 1.3 A
1.5
V  IR
V  1.3 1  1.3 V
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
16

The cell in the diagram has an e.m.f. of 5 V. The current through
the lamp is 0.2 A and the voltmeter reads 3 V. Calculate the internal
resistance of the cell.
E  IR  Ir
IR  t. p.d .  3 V
5  3  0.2  r
r
53
 10 
0.2
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
17

A cell of e.m.f. 4 V is connected to a load resistor of 15 . If 0.2 A
flows round the circuit, what must be the internal resistance of the
circuit?
V  IR
V  0.2 15  3 V
IR  t. p.d .  3 V
Lost volts  4 V  3 V  1 V
Lost volts  Ir
1  0.2  r
r
1
 5
0.2
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
18

A signal generator has an e.m.f. of 8 V and internal resistance of 4
. A load resistor is connected to its terminals and draws a current
of 0.5 A. Calculate the load resistance.
E  IR  Ir
8  0.5  R  0.5  4
R
82
 12 
0.5
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
19

(a) What will be the terminal p.d. across the cell in the circuit below.
t.p.d.  1.5 - 0.2  1.3 V

(b) Will the current increase or decrease as R is increased?
As R increases, the current decreases

(c) Will the terminal p.d. then increase or decrease? Explain your
answer.
The t.p.d. increases because the lost volts will be smaller.
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
20

A cell with e.m.f. 1.5 V and internal resistance 2  is connected to a 3 
resistor. What is the current?
E  I (R  r)
1.5  I (3  2)
I
1.5
 0.3 A
5
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
21

A pupil is given a voltmeter and a torch battery. When he connects the
voltmeter across the terminals of the battery it registers 4.5 V, but when he
connects the battery across a 6  resistor, the voltmeter reading
decreases to 3.0 V.
(a) Calculate the internal resistance of the
battery.
E  4.5 V
t. p.d .  IR  3 V
I 6  3
I  0.5 A
E  IR  Ir
4.5  3  0.5  r
r

4.5  3
 3
0.5
(b) What value of resistor would have to be connected across the battery
to reduce the voltage reading to 2.5 V ?
Lost volts  Ir
2  I 3
I  0.67 A
E  IR  Ir
4.5  0.67  R  2
R
4.5  2
 3.75 
0.67
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
22

In the circuit shown, the cell has an e.m.f. of 6.0 V and internal
resistance of 1 . When the switch is closed, the reading on the
ammeter is 2 A. What is the corresponding reading on the voltmeter
?
Lost volts  Ir
Lost volts  2 1  2 V
E  t. p.d .  Lost volts
t. p.d .  6 V - 2 V  4 V
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
23
In order to find the internal resistance of a cell, the following sets of
results were taken.
 (a) Draw the circuit diagram
used.

Graph of Voltage against Current
1.2
(b) Plot a graph of these results and
from it determine
(i) the e.m.f.
(ii) the internal resistance of the cell.
1
Voltage (V)

0.8
0.6
0.4
0.2
E  IR  Ir
E  Voltage (V )  Ir
V  E  Ir
V  rI  E
E  1.1 V
0
0
0.02
0.04
y = -4.1714x + 1.1053
0.06
0.08
0.1
0.12
0.14
Current (A)
r  4.2 
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
23(cont)

(c) Use the e.m.f. from part (b) to calculate the lost volts for each
set of readings and hence calculate 6 values for the internal
resistance.
E  t. p.d .  Lost volts
Lost volts  E  t. p.d .
Lost volts
Lost volts  Ir

r
Lost volts
I
r ()
0.08 0.16 0.25 0.32 0.41 0.50
4.0 4.0 4.17 4.0 4.10 4.17
(d) Calculate the mean value of internal resistance and the
r
approximate randomMean
uncertainty.
value 
 4.07 
n
Approximat e random uncertaint y 
Maximum value - minimum value
number of measuremen ts
Approximat e random uncertaint y 
4.17 - 4.0
 0.03 
6
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
24

The voltage across a cell is varied and the corresponding current
noted. The results are shown in the table below.
Graph of voltage against current
Plot a graph of V against I.
6.2
Voltage (V)
6.1
6
5.9
5.8
5.7
5.6

(a) What is the open circuit p.d?
Open circuit p.d. is voltage when no current is drawn
5.5
5.4
0
1
2
y = -0.1x + 6
3
4
5
6
Current (A)
E 6V

(b) Calculate the internal
resistance.
E  IR  Ir
E  Voltage (V )  Ir
V  E  Ir
V  rI  E
r  0.1 
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
24 cont

(c) Calculate the short circuit current.
E  I (R  r)
R0
E  Ir
6  I  0.1
I  60 A

(d) A lamp of resistance 1.5  is connected across the terminals of
this supply. Calculate (i) the terminal p.d.and (ii) the power delivered
to the lamp.
E  I (R  r)
6  I (1.5  0.1)
6
I
 3.75 A
1.6
t. p.d  IR  3.75 1.5  5.63 V
P  IV  3.75  5.63  21.1 W
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
25

Calculate the p.d. across R2 in each case.
R2
V2 
 Vs
R1  R2
V2 
8k
5
2k  8k
8
V2   5  4 V
10
R2
V2 
 Vs
R1  R2
V2 
R2
 Vs
R1  R2
1k
5
4k  1k
V2 
750
5
500  750
1
V2   5  1 V
5
V2 
750
5  3 V
1250
V2 
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
26

Calculate the p.d. across AB (voltmeter reading) in each case.
Potential at A = 6 V
Potential at A = 2/7 x 5 = 1.4 V
Potential at A = 2/5 x 10 = 4 V
Potential at B = 3 V
Potential at B = 8/18 x 5 = 2.2 V
Potential at B = 4/10 x 10 = 4 V
p.d. across AB = 3 V
p.d. across AB = - 0.8 V
p.d. across AB = 0
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
27

(a) Calculate the reading on the voltmeter.
Potential at A = 6/15 x 9 = 3.6 V
Potential at B = 3/9 x 9 = 3 V
p.d. across AB = 0.6 V

(b) What alteration could be made to balance the bridge circuit ?
Increase the 9 k resistor to 12 k
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
28

Three pupils are asked to construct balanced Wheatstone bridges.
Their attempts are shown.
One of the circuits gives a
balanced Wheatstone
bridge, one gives an off balance Wheatstone bridge
and one is not a
Wheatstone bridge.
 (a) Identify each circuit. Unbalanced WheatstoneBalanced WheatstoneNon Wheatstone
 (b) How would you test that balance had been obtained?

The galvo should read zero in a balanced Wheatstone Bridge
(c) In the off – balance Wheatstone bridge (i) calculate the potential
difference across the galvanometer. Potential at A = 5/15 x 1.5 = 0.5 V
Potential at B = 10/15 x 1.5 = 1 V
 (ii) in which direction will electric
p.d. across AB = 0.5 V
current flow through the
galvanometer?
From B to A
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
Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
29

Calculate the value of the unknown resistor X in each case.
R1 R3

R2 R4
R1 R3

R2 R4
120 120

X
9
4k 12k

15k
X
X 9
12k 15k
X
 45 k
4k
R1 R3

R2 R4
10k 3.6k

25k
X
X
3.6k  25k
 9 k
10k
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Physics Support Materials
Higher
Electricity and Electronics
Electric Fields and Resistors in Circuits
30
The circuit shown opposite is balanced.


(a)What is the value of resistance X?
R1 R3

R2 R4

10 5

20 X
X  10 
(b) Will the bridge be unbalanced if
(i) a 5  resistor is inserted next to the 10  resistor
(ii) a 3 V supply is used.?The bridge will be unbalanced if the resistance is increased to 15
The bridge will still be balanced if the supply voltage is changed

(c) What is the function of resistor R and what is the disadvantage
of using it as shown?
The resistor R protects the sensitive galvo from large currents. In series the
resistor reduces the sensitivity of the galvo
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