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A sinusoidal ac current flows through a resistor of resistance R . If the peak current is `I_(p)` , then the power dissipated isA. `I_(1)^(p)R cos theta`B. `1/2I_(p)^(2)R`C. `4/pi I_(p)^(2)R`D. `1/pi^(2)I_(p)^(2)R` |
Answer» Correct Answer - B `ltP gt = I_(rms)^(2) R=(I_(p)/sqrt2)^(2) R=(I_(p)^(2)R)/2` |
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