Skip to main content
GIOPHYSICS
Menu

IGCSE · 4 · structured

Electricity and magnetism · Question 7

Electricity and magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
10
Topics
2
Answer
Complete
A 40 ohm resistor in series with a parallel pair of 30 ohm and 60 ohm resistors40 Ω30 Ω60 Ω12 V
Fig. 7.1Circuit diagram. From the positive terminal of the 12 V battery, drawn at the foot of the circuit with its long plate on the left, the wire runs round to a resistor labelled 40 ohm. Beyond that resistor the circuit reaches a junction and divides into two parallel branches, the upper one containing a resistor labelled 30 ohm and the lower one a resistor labelled 60 ohm. The branches rejoin at a second junction, and a single wire returns from there to the negative terminal of the battery.
structured10 marks

A 12 V battery of negligible internal resistance is connected to a 40 Ω resistor in series with a parallel combination of a 30 Ω resistor and a 60 Ω resistor.

  1. (a)

    Calculate Calculate the combined resistance of the parallel pair.

    2 marks
  2. (b)

    Calculate Calculate the current drawn from the battery.

    2 marks
  3. (c)

    Determine Determine the potential difference across the parallel pair.

    2 marks
  4. (d)

    Determine Determine the current in the 60 Ω resistor.

    2 marks
  5. (e)

    Explain The 60 Ω resistor is now removed from the circuit, leaving a break in that branch. Explain what happens to the current drawn from the battery.

    2 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: 1/R = 1/30 + 1/60 = 3/60 R = 20 Ω total resistance = 40 + 20 = 60 Ω
01

(a)

2 marks

Calculate Calculate the combined resistance of the parallel pair.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1/R = 1/30 + 1/60 = 3/60

  2. 2

    R = 20 Ω

02

(b)

2 marks

Calculate Calculate the current drawn from the battery.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    total resistance = 40 + 20 = 60 Ω

  2. 2

    I = V / R = 12 / 60 = 0.20 A

03

(c)

2 marks

Determine Determine the potential difference across the parallel pair.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    p.d. across the 40 Ω resistor = 0.20 × 40 = 8.0 V

  2. 2

    p.d. across the pair = 12 − 8.0 = 4.0 V (or 0.20 × 20 = 4.0 V)

04

(d)

2 marks

Determine Determine the current in the 60 Ω resistor.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    I = V / R = 4.0 / 60

  2. 2

    I = 0.067 A (67 mA)

05

(e)

2 marks

Explain The 60 Ω resistor is now removed from the circuit, leaving a break in that branch. Explain what happens to the current drawn from the battery.

How to approach it

State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.

  1. 1

    the total resistance increases, because the parallel pair is replaced by the 30 Ω resistor alone, giving 70 Ω instead of 60 Ω

  2. 2

    so the current from the battery decreases (to about 0.17 A)

Private on this device

Did your answer earn the marks?

Compare the reasoning point by point, then choose what happens next.