This document provides answers to questions about electric circuits from a student book. 1) It defines key terms like current, potential difference, resistance, and charge. 2) It explains how to calculate values in series and parallel circuits such as total resistance, current, and potential difference across components. 3) Sample calculations are provided for circuits involving combinations of cells, resistors, ammeters, and voltmeters.

1. P4 Electric circuitsStudent Book answers
P4.1 Electrical charges and fields
Question
number
Answer Marks Guidance
1 a i electrons transfer from cloth to polythene rod 1
1 a ii electrons transfer from perspex rod to cloth 1
1 b electrons negatively charged,
glass loses electrons so gains positive charge
1
1
2 a attract 1
2 b repel 1
3 a attract 1
3 b attract 1
3 c repel 1
4 a X and Y have same type of charge 1
4 b Suspend R horizontally, rub with dry cloth to charge, charge X and hold
near R,
if X repels R, X also negative,
if X attracts R, X positive,
Y repels X so same charge as X
1
1
1
1
5 a friction between soles of shoes and carpet causes you and shoes to
become charged,
touching metal radiator gives shock because radiator earthed and you are
charged at very high voltage,
electrons transfer between you and radiator as a spark
1
1
1
5 b as you move in seat clothing rubs against car seat fabric so you become
charged,
if metal frame of car earthed,
when you get out of seat you get electric shock when touch metal frame of
car because charge on you creates a spark between you and car frame
1
1
1
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2. P4 Electric circuitsStudent Book answers
P4.2 Current and charge
Question
number
Answer Marks Guidance
1 1 = cell,
2 = switch,
3 = indicator,
4 = fuse
2 3 correct gains 1 mark
2 a circuit correct: diode at 2 with arrow pointing to 3 1
2 b variable resistor 1
2 c 0.25 A × 60 s
= 15 C
1
1
3 a measure electric current 1
3 b change current in the circuit 1
4 a circuit with bulb, wires and cell 2 one component incorrect gains 1 mark
4 b electron passing through battery gains energy from chemical reactions
in battery,
electron transfers energy to filament bulb by colliding with atoms in
filament as it passes through,
transfers some energy to atoms in wire in same way
1
1
1
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3. P4 Electric circuitsStudent Book answers
P4.3 Potential difference and resistance
Question
number
Answer Marks Guidance
1 a
= 8.0 Ω 1
1 b suitable values read off graph and used
to give 10.0 Ω
1
1
2 W 6.0 Ω
X 80 V
Y 2.0 A
1
1
1
3 a
= 800 Ω 1
3 b i 0.015 A × 1200 s = 18 C 1
3 b ii 18 C × 12.0 V = 216 J 1
4 a suitable values read off graph and used
to give 10.0 Ω
1
1
4 b i
= 0.16 A
1
1
4 b ii 0.42 A × 10.0 Ω
= 4.2 V
1
1
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4. P4 Electric circuitsStudent Book answers
P4.4 Component characteristics
Question
number
Answer Marks Guidance
1 a i thermistor 1
1 a ii diode 1
1 a iii filament bulb 1
1 b i
= 5 Ω 1
1 b ii
= 10 Ω 1
2 a
= 15 Ω 1
2 b ammeter reading increases,
because resistance of thermistor decreases,
so total resistance decreases
1
1
1
3 if LDR covered current decreases
as LDR resistance increases
and p.d. still 9.0 V
1
1
1
4 a current = 0 until p.d. ≈ 0.7 V
then increases rapidly
1
1
4 b resistance very large until ≈ 0.7 V
then decreases rapidly
1
1
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5. P4 Electric circuitsStudent Book answers
P4.5 Series circuits
Question
number
Answer Marks Guidance
1a 1.2 V – 0.8 V
= 0.4 V
1
1
1 b
I = = 0.20 A,
p.d. = 1.5 V – 1.0 V = 0.5 V
1
1
2 a circuit with cell and 2 resistors in series correct 1
2 b i 3.0 Ω + 2.0 Ω = 5.0 Ω 1
2 b ii
= 0.3 A
1
1
2 c
total R = = 6.0 Ω
RX = 6.0 Ω – 2.0 Ω = 4.0 Ω
1
1
3 a i 2 Ω + 10 Ω = 12 Ω 1
3 a ii 2 × 1.5 V = 3.0 V 1
3 b
= 0.25 A 1
3 c VP = 0.25 A × 2 Ω = 0.5 V
VQ = 0.25 A × 10 Ω = 2.5 V
1
1
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6. P4 Electric circuitsStudent Book answers
Question
number
Answer Marks Guidance
3 d i 2 Ω + 10 Ω + 5 Ω = 15 Ω 1
3 d ii
= 0.20 A
1
1
3 d iii VP = 0.20 A × 2 Ω = 0.4 V
VQ =0.20 A × 10 Ω = 2.0 V
VR =0.20 A × 3 Ω = 0.6 V
1
1
1
4 any four from:
• same current through each resistor,
• with additional resistor in series more resistors share total p.d. ,
• so p.d. across each resistor less,
• current through resistors less,
• total p.d. unchanged so total resistance 





=
current
p.d.total
> before
4
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7. P4 Electric circuitsStudent Book answers
P4.6 Parallel circuits
Question
number
Answer Marks Guidance
1 a 0.40 A − 0.10 A = 0.30 A 1
1 b 3 Ω resistor 1
1 c
battery current = 10 A ∴ current same if R of single resistor =
= 0.60 Ω
1
1
2 a circuit diagram: 6.0 V battery across 12 Ω and 24 Ω resistors in parallel 1
2 b i
current = = 0.50 A 1
2 b ii
current = = 0.25 A 1
2 c cell current = 0.5 A + 0.25 A
= 0.75 A
1
1
3 a i
I1 = = 3.0 A
I2 = = 2.0 A
I3 = = 1.0 A
1
1
1
3 a ii 6.0 A 1
3 b
I through R3 = 1.5 A
total I = 3.0 A + 2.0 A + 1.5 A = 6.5 A
1
1
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8. P4 Electric circuitsStudent Book answers
Question
number
Answer Marks Guidance
4 I through 2 Ω resistor = 3.0 A
total I = sum of currents in individual resistors so > 3.0 A
equivalent R =
as total current > 3.0 A, equivalent R < 2 Ω
(or current through 2 Ω resistor < total current
as all resistors contribute to total current,
p.d. across 2 Ω resistor = p.d. across combination,
so equivalent R < 2 Ω as total current > current through 2 Ω resistor and
p.d. is the same)
1
1
1
1
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