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The potential differences `V` and the current `i` flowing through an instrument in an `AC` circuit of frequency `f` are given by `V=5 cos omega t` and `I=2 sin omega t` amperes (where `omega=2 pi f`). The power dissipated in the instrument isA. zeroB. `10W`C. `5W`D. `2.5W` |
Answer» Correct Answer - A `V =5 cos omega t = 5 sin (omega t + pi//2)` `I = 2 sin omegat` phase different between `V` and I `phi = pi//2` `P =0` . |
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