IGCSE · Physics 0625 · Electricity & magnetism

Series & parallel circuits: IGCSE Physics questions with answers

Free Series & parallel circuits questions for IGCSE Physics (0625), part of Electricity & magnetism. Each question has the answer and the marking points an examiner would look for.

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Series & parallel circuits practice questions

Question 1Multiple choice · 1 markA* challenge

The battery has negligible resistance.
What is the current in the 6.0 Ω resistor?

3.0 Ω6.0 Ω6.0 Ω6.0 V
Fig. 1.1
  1. A 0.25 A
  2. B 0.40 A
  3. C 0.75 A
  4. D 1.0 A
Show answer

Answer: A

Question 2Multiple choice · 1 markA* challenge

Two identical lamps X and Y are connected in series with a battery. A third identical lamp Z is then connected in parallel with lamp Y, as shown.
How do the brightnesses of X and Y change when Z is connected?

YZX
Fig. 2.1
lamp Xlamp Y
Adimmerdimmer
Bno changedimmer
Cbrighterbrighter
Dbrighterdimmer
Show answer

Answer: D

Question 3Multiple choice · 1 mark

How is a voltmeter connected to measure the p.d. across a lamp?

  1. A in series with the lamp
  2. B in series with the battery only
  3. C between the switch and the ammeter
  4. D in parallel with the lamp
Show answer

Answer: D

Question 4Structured · 4 marks

A 2.0 Ω resistor and a 3.0 Ω resistor are connected in series with a 10 V supply.

(a)

Calculate the current.

[2]
(b)

Calculate the potential difference across the 3.0 Ω resistor.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)2.0 A
R = 5.0 Ω C1
I = 2.0 A A1
2
(b)6.0 V
V = IR = 2.0 × 3.0 C1
6.0 V A1
2
Question 5Structured · 9 marks

A 6.0 V battery is connected to a 2 Ω resistor in series with two resistors of 6 Ω and 24 Ω connected in parallel with each other.

(a)

Calculate the combined resistance of the two resistors in parallel.

[2]
(b)

Use your answer to (a) to calculate the current in the battery.

[2]
(c)

Calculate the potential difference across the 2 Ω resistor.

[1]
(d)

Calculate the current in the 6 Ω resistor.

[2]
(e)

State two advantages of connecting lamps in parallel in a house.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)4.80 Ω
1/R = 1/6 + 1/24 C1
4.80 Ω A1
2
(b)0.882 A
total R = 4.80 + 2 = 6.80 Ω C1
I = 6.0/6.80 = 0.882 A A1
2
(c)1.76 V
V = IR = 0.882 × 2 = 1.76 V B1
1
(d)0.706 A
p.d. across parallel part = 4.24 V C1
÷ 6 = 0.706 A A1
2
(e)each can be switched independently; if one fails others stay on; each gets full p.d.
Any 2 of: lamps can be switched on and off separately; if one lamp breaks the others still work; each lamp has the full supply p.d. B2
2
Question 6Structured · 7 marks

Fig. 6.1 shows two resistors connected in parallel to a battery.

4.0 Ω12.0 ΩA12.0 V
Fig. 6.1
(a)

Calculate the combined resistance of the two resistors.

[2]
(b)

Calculate the reading on the ammeter.

[2]
(c)

Calculate the current in the 12.0 Ω resistor and the power dissipated in it.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)3 Ω
1/R = 1/R1 + 1/R2 C1
3 A1
2
(b)4 A
I = V ÷ R C1
4 A1
2
(c)1 A, 12 W
I = 12 ÷ 12 C1
1 A1
P = IV = 12 A1
3

More Electricity & magnetism questions

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in