Electricity
5 marks · Normal3. A string of decorative garden lights contains many bulbs connected in parallel to the mains supply. Each bulb is marked '230 V, 15 W'.
Explain why the bulbs are connected in parallel rather than in series.
Calculate the current in a single bulb.
If all 40 bulbs are switched on, calculate the total current drawn from the supply.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Parallel connection ensures each bulb receives full voltage, so each shines with normal brightness; if one bulb fails, others remain lit, unlike series where all go out and bulbs are dimmer. · must show voltage same for each bulb, normal brightness, fault does not affect all bulbs
(b) 0.065 A · must show I=P/V, 15W/230V, =0.065A
(c) 2.6 A · must show multiply I of one bulb by 40, 40×0.065=2.6
Common mistakes: Incorrectly using series calculation; Power/current formula confusion
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States that parallel gives each bulb full voltage/same brightness.
B2 (a) · 1 mark · Mentions that if one bulb fails, others stay lit (or series: all go out).
M1 (b) · 1 mark · Uses I = P/V for one bulb.
A1 (b) · 1 mark · Correct value 0.065 A or rounded (accept 0.0652), with or without work. (follow-through)
A2 (c) · 1 mark · Multiplies single bulb current by 40: 2.6 A. (follow-through)
Worked solution
a. Bulbs are connected in parallel so each receives the full mains voltage (230 V). This means each shines at normal brightness. If a bulb fails, the others keep working, unlike a series circuit.
b. I = P / V = 15 W / 230 V = 0.065 A (rounded to 2 decimals: 0.065 A).
c. Total current = 40 × 0.065 A = 2.6 A.