CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Coursebook answers
Chapter 11
Rearrange e.m.f. E = IR + Ir to give internal
resistance of battery
Self-assessment questions
1
E = 5.0 V
r = (E − IR)/I = (3.0 − 2.8)/0.28
= 0.71Ω
r = 2.0 Ω
current = 0.50 A
5
1.5
1.0
R = 8.0 Ω
V
0.5
Rearrange e.m.f. E = I(R + r) to give
current
I = E/(R + r) = 5.0/(8.0 + 2.0) = 0.50 A
2
a
i Rearrange e.m.f. E = I(R + r) to give
current
I = E/(R + r) = 3.0/(10 + 10 + 4.0) =
0.125 ≈ 0.13 A
0
0.0
6
1/R = 1/R1 + 1/R2 = 1/10 + 1/10,
so R = 5.0 Ω
b
3
4
c
0.8
1.0
power, P = VI, and p.d. V = IR means
current, I = V/R, so P = V2/R
power, P = V2/R = 82/4 = 16 W
7
hen resistor is set to 0 Ω, Vout = 0 V
W
When resistor is set to 40 Ω, Vout = R2/(R1 + R2)
× Vin = 40/(10 + 40) × 10 = 8 V
8
rom the graph in Figure 12.7, the resistance
F
of the LDR is 100 kΩ.
voltage V across the 3 kΩ = R1/(R1 + R2) × Vin,
where R1 is the resistance of the
Rearrange e.m.f. E = I(R + r)
to give current I = E/(R + r) = (1.5 × 4)/(2.0 +
(0.1 × 4)) = 2.5 A
E = 3.0 V
p.d. across 10 Ω resistor = 2.8 V so current in
circuit with resistor connected
0.6
Rearrange to give resistance R = V2/P =
(12)2/36 = 4.0 Ω
iilost volts = Ir = 0.33 × 4.0 = 1.33 V
terminal p.d. = e.m.f. − lost volts = 3.0
− 1.33 = 1.67 V
I/A
a
terminal p.d. = E − Ir = 12 − (100 × 0.04)
=8V
b
ilost volts = Ir = 0.125 × 4.0 = 0.5 V
terminal p.d. = e.m.f. − lost volts = 3.0
− 0.5 = 2.5 V
0.4
E = 1.5 V, r = 0.5 Ω
iiExternal resistance R is given by
so, current I = E/(R + r) = 3.0/(5.0 +
4.0) = 0.33 A
0.2
3 kΩ resistor and R2 is the resistance of the
LDR
V = 3/(3 +100) × 10 = 0.29 V
9
Connect the output across the 3 kΩ resistor.
= V/R = 2.8/10 = 0.28 A
1
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
10 T
he thermistor is connected in series with
a fixed resistor and a battery. A changing
temperature will cause a changing voltage
across the thermistor.
11 Both are made from a semiconductor
material. Both have a decreasing resistance:
for an LDR when the light intensity increases
and for a thermistor when the temperature
increases. Both have a non-linear change in
resistance with light intensity or temperature.
12 The resistance of the thermistor at 50° C is
400 Ω so the resistor must have the same value.
13 a
For a 1 cm length of wire, potential
difference = 4.0/100 = 0.04 V
length needed for 1.0 V = 1.0/0.04 = 25 cm
b
Vout
c
he output voltage is shown across the 400
T
Ω resistor. When the temperature rises, the
2
length of 37.0 cm has a p.d. across it of
A
37.0 × 0.04 = 1.48 V
The driver cell will have internal resistance
and it is supplying current to the
potentiometer wire. Therefore, the p.d.
across its terminals and the wire will be
slightly less than the e.m.f. (4.0 V) of the
cell.
10 V
400 Ω
resistance of the thermistor decreases and so
the p.d. across the thermistor decreases and
the p.d. across the 400 Ω resistor increases.
You can instead put the output voltage across
the thermistor, then, when temperature rises,
the output voltage falls.
I f a balance length of 31.2 cm is required
by a cell of e.m.f. 1.230 V, then p.d.
supplied by unknown e.m.f. cell = (1.230 ×
37.0)/31.2 = 1.459 V ≈ 1.46 V
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020