Home / iGCSE Physics (0625) 4.2.4 Resistance Paper 3 -Exam Style Questions- New Syllabus

iGCSE Physics (0625) 4.2.4 Resistance Paper 3 -Exam Style Questions- New Syllabus

Question

Fig. 8.1 shows part of a circuit for measuring the resistance of a lamp.
(a) Draw on Fig. 8.1 to show how to connect a voltmeter to measure the potential difference across the lamp. Use the electrical symbol for a voltmeter.
(b) The current in the lamp is $0.41\text{ A}$ and the potential difference across the lamp is $12\text{ V}$. Calculate the resistance of the lamp.
(c) Calculate the electrical power transferred in the lamp. Include the unit.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

Topic 4.2.4 — Resistance (Part (b))
Topic 4.2.5 — Electrical energy and electrical power (Part (c))
Topic 4.3.1 — Circuit diagrams and circuit components (Part (a))

▶️ Answer/Explanation

(a)
For the correct answer:
correct symbol voltmeter in parallel with lamp

A voltmeter must be connected in parallel with the component across which the potential difference is being measured. The symbol for a voltmeter is a circle with a ‘V’ inside, and it should be drawn with leads connected to the two terminals of the lamp.

(b)
For the correct answer:
$29\ \Omega$

The resistance is calculated using the equation $R = \frac{V}{I}$. Substituting the given values, $R = \frac{12\text{ V}}{0.41\text{ A}}$, which yields approximately $29.3\ \Omega$. Rounded to two significant figures to match the precision of the given current, the result is $29\ \Omega$.

(c)
For the correct answer:
$4.9\ \text{W}$

Electrical power is calculated using $P = I \times V$. Substituting the values, $P = 0.41\text{ A} \times 12\text{ V} = 4.92\text{ W}$. Rounded to two significant figures, the power transferred is $4.9\text{ W}$.

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