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The output of a step-down transformer is measured to be 24 V when connected to a 12 W light bulb. The value of the peak current (Iₘ) is _____.

Asked in GUJCET 2023 · RMS and mean values

Answer: (2) 0.71 A

Step-by-step solution

Given: the measured output 24 V is an r.m.s. value, and the bulb takes P=12 W.

Idea: for a resistive load P=VᵣₘₛIᵣₘₛ.

Iᵣₘₛ=P/(Vᵣₘₛ)=(12)/(24)=0.5 A.

Peak current: Iₘ=√2 Iᵣₘₛ=1.414×0.5=0.707 A.

So Iₘ≈0.71 A, that is 1/(√2) A.

Why the other options are wrong

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