Home / iGCSE Physics (0625) 4.3.2 Series and parallel circuits Paper 3 -Exam Style Questions- New Syllabus

iGCSE Physics (0625) 4.3.2 Series and parallel circuits Paper 3 -Exam Style Questions- New Syllabus

Question

A battery, a lamp L, a fixed resistor R and a switch S are connected as shown in Fig. 7.1.

(a) The potential difference (p.d.) across lamp L is 4.8V and the current in lamp L is 0.40A. Calculate the resistance of lamp L.

(b) State and explain how closing switch S affects the brightness of lamp L.

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

Topic 4.2.4 — Resistance (Part (a))
Topic 4.3.2 — Series and parallel circuits (Part (b))

▶️ Answer/Explanation

(a)
12 \( \Omega \)
Resistance is calculated using the equation \( R = \frac{V}{I} \). Substituting the given values: \( R = \frac{4.8\text{ V}}{0.40\text{ A}} = 12\ \Omega \).

(b)
Lamp L becomes brighter (brightness increases).
Closing switch S places resistor R in parallel with the wire, significantly reducing the total resistance of that branch. This decreases the overall circuit resistance, increasing the total current from the battery and the potential difference across lamp L, thus causing a higher current through the lamp.

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