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
5
Exam-style questions
1
B[1]
2
C[1]
3
a
terminal p.d. = 2.5 × 0.30 = 0.75 V [1]
which equals the ‘lost volts’ divided by
the current.[1]
There is work done inside the cell against
the internal resistance. or There is a
voltage (lost volts) across the internal
resistance.[1]
b
a
The resistance due to the work done (or
energy transferred) in driving current
through the cell[1]
b
r = 0.40 Ω[1]
E = V + Ir [1]
iiSubstitution into E = I(R + r) so
E = 1.50 V[1]
c
so, r = 0.30 Ω [1]
power P = I 2 R = 2.52 × 0.30
[1]
= 1.875 ≈ 1.88 W
iiPower for 0.5 Ω: total resistance
R + r = 0.80 Ω
1.5
current
= = 1.875 A [1]
0.8
i
6
aThe e.m.f. of a cell is the work done per
coulomb of charge [1] in the complete
circuit.[1]
b
Power for 0.2 Ω: total resistance
R + r = 0.50 Ω
1.5
current
= = 3.0 A [1]
0.5
power = 3.02 × 0.20 = 1.80 W [1]
c
a
i The test cell is the wrong way round
[1]
iiAt the balance point, the ammeter
reading is zero.[1]
1
e.m.f . 2.25
[1]
=
1.434 34.6
2.25
so, e.m.f . =
× 1.434 = 0.933 V [1]
34.6
here is no/negligible current through
T
the high resistance voltmeter and, hence,
the cell. When the resistor is connected
in parallel there is a much larger current
through that and the cell.[1]
There is now a potential drop as electrical
work is done against the internal
resistance of the cell.[1]
V 8.40
i I= =
[1]
R 12
= 0.70 A[1]
iilost volts = 0.54 V[1]
lost volts 0.54
r=
=
= 0.77 Ω[1]
I
0.7
iiiThe resistance of the voltmeter >> r
or R[1]
so he must reverse it.[1]
b
Internal resistance is too high[1]
Maximum current < 4 A[1]
power = 1.8752 × 0.5 = 1.76 W [1]
4
E = I ( R + r ) = 0.625 ( 2 + r )[1]
= 0.341( 4 + r )
1.5 = 0.75 + 2.5 × r so 2.5r = 0.75[1]
c
i
7
a
In circuit 1, the p.d. across the bulb
varies from 0 to 240 V.[1]
In circuit 2, it never falls to zero.[1]
V 2 2402
=
R =
b i
[1]
60
P
= 960 Ω[1]
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
iiResistance is greater when switched
on. or Resistance is lower at room
temperature.[1]
Resistance of a metal increases with
increasing temperature (or decreases
with decreasing temperature).[1]
8
a
Voltmeter reading will decrease[1]
because current through R2 decreases.[1]
R2
470
×Vin so, 2.0 =
×9
b Vout =
( R1 + R2 )
470 + R1
[1]
R1 = 1645 ≈ 1600 Ω[1]
c
resistance of R2 and voltmeter in parallel
−1
 1   1 
= 
  = 903 Ω
+
  1645   2000 )  
Vout =
9
[1]
903
R2
×Vin =
= 5.9V [1]
( R1 + R2 )
(1645 + 903)
a
i Straight line through origin with
positive gradient[1]
Graph axes labelled V (x-axis) and
l (y-axis)[1]
b
E
= I ( R + r )[1]
= 0.6 × ( 8 + r ) = 1.50 × ( 2 + r )[1]
r = 2.0 Ω[1]
Substitution into either equation gives
E = 6.0 V[1]
10 a
A diagram similar to Figure 11.13[2]
b
i± 0.2 cm[1]
R1
15.4
ii
=
= 0.36[1]
( R1 + R2 ) 42.6
iii
R1
15.4
[1]
=
R2 ( 42.6 − 15.4 )
= 0.57 [1]
ivuncertainty in R1 = ±0.2 cm;
percentage uncertainty = 1.3%
uncertainty in R2 = ± 0.2 cm;
percentage uncertainty = 0.5%[1]
total percentage uncertainty
= 10.7 + 0.5 = 11.2 ≈ 11%
total uncertainty = ±0.06[1]
iiA: 0 V; B: 2.2 V[1]
iii General diagram (with one or two
cells)[1]
Two cells in correct polarity[1]
Switches, or suitable comment
indicating that only one cell is used
at a time[1]
2
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020