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The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is [NCERT Exemplar] (a) 1/√2 A. (b) √2A. (c) 2 A. (d) 2√2 A. |
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Answer» (a) 1/√2Explanation:GIVEN:P=12 and V=24Find: I• PEAK CURRENT Now,P=VI I=P/V I=12/24=1/2Then Irms=I•/√2 I•=Irms×√2 I•=1/2×√2 I•=1/√2A. THANKYOU |
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