Module 3 · Answers
Answers & explanations
Workings shown for the calculation questions.
Section A — Easy
01
(a) ampere (A) (b) volt (V) (c) ohm (Ω)
02
V = I × R (or R = V/I)
03
(a) In series with the component (b) In parallel across the component
04
Iron, nickel, cobalt (also steel).
05
Any two: more turns of wire, larger current, soft iron core.
Section B — Medium
06
57.5 Ω
R = V ÷ I = 230 ÷ 4 = 57.5.
07
2 A
I = V ÷ R = 12 ÷ 6.
08
2 V across each.
In series, the battery voltage is shared equally between identical components: 6 ÷ 3 = 2 V.
09
A series circuit is a single loop. A broken bulb breaks the loop, so no current can flow anywhere — every bulb goes out.
10
Rubbing transfers electrons between balloon and jumper, leaving the balloon charged. The charged balloon attracts uncharged paper because it induces opposite charges in the paper.
11
An electromagnet can be switched on (to pick up) and off (to release) by turning the current on or off. A permanent magnet would always attract — you'd never be able to drop the load.
Section C — Hard
12
(a) 8 Ω (b) 1.5 A (c) 6 V across each
In series, total R = 4 + 4 = 8.
I = 12 ÷ 8 = 1.5 A.
V across each = I × R = 1.5 × 4 = 6 V.
I = 12 ÷ 8 = 1.5 A.
V across each = I × R = 1.5 × 4 = 6 V.
13
12 V across each. 3 A in each branch. Total current from battery = 6 A.
In parallel, each branch has the full 12 V.
Branch current: I = V ÷ R = 12 ÷ 4 = 3 A.
Branch current: I = V ÷ R = 12 ÷ 4 = 3 A.
14
Independent: number of turns. Dependent: e.g. number of paper clips picked up. Controls: same current, same battery, same iron core, same coil length.
Method: build the electromagnet, count paperclips picked up. Repeat with different turn counts (e.g. 10, 20, 30, 40), keeping everything else the same.
Method: build the electromagnet, count paperclips picked up. Repeat with different turn counts (e.g. 10, 20, 30, 40), keeping everything else the same.
15
A bulb in the parallel circuit is brighter. In the parallel circuit each bulb gets the full 6 V; in the series circuit each gets 3 V (shared). More voltage across a bulb means more current through it and more energy transferred per second, so it's brighter.