Cambridge International AS Level Physics Answers to self-assessment questions

Answers to SAQs
Chapter 9 11 Current I = nAve, but remember that, in a
series circuit, current is the same all the
1 Towards the right. way round the circuit. Also, A and e are the
same for both lengths of wire. This means
a
2 a, b, c A – + B that n1Av1e = n2Av2e, where the 1 denotes the
c copper wire and the 2 denotes the silver wire.
V n 28
c c So that means V2 = n1 = 8.5 × 1028 = 1.4.
1 2 5.9 × 10
+ This means that the drift velocity in the
– silver wire is 1.4 times the drift velocity in the
copper wire.
b
12 Current I = RV = 12
36 = 0.33 A
3 Charge ΔQ = IΔt = 0.4 × 15 = 6.0 C
13 A 60 W lamp has higher resistance. It allows
4 Rearrange ΔQ = IΔt so I = ΔQ 150
Δt = 30 = 5 A
less current through for the same potential
difference (power is current × potential
5 Rearrange ΔQ = IΔt so I = ΔQ 50
Δt = 20 = 2.5 A difference, so, applying Ohm’s law, power is
inversely proportional to resistance).
6 a
Rearrange ΔQ = IΔt and use 50 A per hour as
the charge available, so 14

a Potential difference, V = IR = 1.0 × 50 = 50 V
Δt = ΔQ 50
I = 200 = 0.25 hours = 15 minutes b Potential difference, V = IR = 2.0 × 50 = 100 V
b ΔQ = IΔt; for a current of 200 A, 15 230 = 580 Ω (to 2 sig. figs)
Resistance R = VI = 0.40
ΔQ = 200 × 15 × 60 = 180 000 C
16 Resistance R = VI = 6.0
2.4 = 2.5 Ω
7 Number of protons in 1 C
= 1 18
17 Power P = VI, so rearrange to give current:
–19 = 6.2 × 10 protons
1.6 × 10 60
I = PV = 230 = 0.26 A
8 The smallest possible unit of electric charge
is the elementary charge, 1.6 × 10–19 C. So the 18 Power P = VI = 25 × 103 × 40 × 103 = 1.0 × 109 W
only possible electrical charges are whole (1000 MW or 1 GW)
multiples of this elementary charge. The only
value from the list that is a whole multiple is 19 a Power P =4 VI so rearrange to give current:
8.0 × 10–19 C. 10
I = PV = 230 = 43 A
9 Current I = nAve b Fuse needs to be rated above, but close to,
28 –6 –3
= 5.9 × 10  × 2.0 × 10  × 0.1 × 10  × 1.6 × 10–19 full normal current draw, so a fuse of 45 to
= 1.9 A 50 A would be appropriate.
2 2

10 Cross-sectional area of copper wire 20 Power P = VR = (3.0) 3 = 0.45 mW

2 –3 2
(20 × 10 )
πd
2 π × (1.0 × 10 )
A = πr = 4 = 4 = 0.79 × 10–6 m2 21
P 15
a Current I = V = 230 = 0.065 A
Rearrange I = nAve to give drift velocity 230
b Resistance R = VI = 0.065 = 3500 Ω
I =
v = nAe 5.0
28 –6 –19
8.5 × 10  × 0.79 × 10  × 1.6 × 10
–3 –1 –1 22 Rearrange power P = I2R to give resistance:
= 0.47 × 10  m s = 0.47 mm s
R = P2 = 100 2 = 540 Ω
I (0.43)

Cambridge International AS and A Level Physics © Cambridge University Press 2014

 Cambridge International AS Level Physics Answers to self-assessment questions

23 Energy transferred
W = IVΔt = 10 × 12 × (5.0 × 60 × 60)
= 2200 000 J, or 2.2 MJ

24

a Charge through lamp
ΔQ = IΔt = 10 × 20 = 200 C
b Per coulomb of charge,
W 400
energy = ΔQ = 200 = 2.0 J C–1
c Rearrange energy transferred, W = IVΔt,
to give potential difference: V = ΔW
IΔt
400
= (10 × 20) = 2.0 V
energy
or use potential difference = charge
= 400
200
= 2.0 V

Cambridge International AS and A Level Physics © Cambridge University Press 2014