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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. 2AD. `2sqrt(2)`A |
Answer» Correct Answer - A secondary voltage `V_(S) = 24V` Power associated with secondary `P_(S) =12 W` `I_(s)=P_(S)/V_(s)=12/24` `=1/2A=0.5A` Peak value of the current in the secondary `I_(0) = I_(S) sqrt(2)` `=(0.5)(1.414)=0.707=1/sqrt(2)A` |
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