WAEC 2015 Physics · Question 41

An alternating current supply is connected to an electric lamp which lights with the same brightness as it…

Question

An alternating current supply is connected to an electric lamp which lights with the same brightness as it does with direct current source emf 6v. The peak potential difference of the a.c supply is

A.4.6v
B.6.0v
C.8.5v
D.12.0v

Answer: C. 8.5v

Explanation written with AI and checked against the answer key

Analyze the question The lamp has the same brightness on AC as on a 6 V DC source. Same brightness means the same power, so the effective (r.m.s.) value of the AC voltage equals the DC voltage: V_rms = 6 V. We are asked for the peak voltage. Concept For a sinusoidal AC supply: V_rms = V₀ / √2, so V₀ = √2 × V_rms Solution V₀ = √2 × 6 V₀ = 1.414 × 6 V₀ ≈ 8.49 V ≈ 8.5 V Answer The peak potential difference is about 8. Tip "Same brightness as DC" or "same heating effect" always means r.m.s. = DC value. Then peak = r.m.s. × 1.414. Don't pick 6.0 V, because that is the r.m.s. value, not the peak. Peak is always larger than r.m.s.

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