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If $I=I_0\sin(ωt-\frac{π}{2})$ and $E = E_0 \sin ωt$, then the AC power consumed is |
$P=\frac{E_0I_0}{\sqrt{2}}$ $P=\sqrt{2}E_0I_0$ $P=\frac{E_0I_0}{2}$ $P=0$ |
$P=0$ |
The correct answer is Option (4) → $P=0$ The average power consumed by the circuit is zero because the current and the voltage are out of phase by 90°, meaning that no real power is transferred to the load. |
