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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 isA. `(1)/(sqrt(2))A`B. `sqrt(2)A`C. `2A`D. `2sqrt(2)A` |
Answer» Correct Answer - A Here, `V_(s)=24 V, P_(s) =12 W` `I_(s)=(P_(s))/(V_(s))=(12)/(24)=0.5 A , I_(m)=sqrt(2) I_(s)=sqrt(2)xx0.5=(1)/(sqrt(2)) A` |
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