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Electric Power

The potential...

Question

The potential difference $V$ and the current $I$ flowing through an instrument in an AC circuit are given by $V = 5 cos ω t v o l t, I = 2 sin ω t A m p$ . The power dissipated in the circuit is

5

watt

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10

watt

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2.5

watt

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0

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Solution

The correct option is D $0$
The voltage can be written as,

$V = 5 (sin ω t + \frac{π}{2})$

The angle between the voltage and the current is $90^{\circ}$ .

The power dissipated in the circuit is given as,

$P = V_{r m s} I_{r m s} cos ϕ$

$= V_{r m s} I_{r m s} cos 90^{\circ}$

= $0$

Thus, the power dissipated in the circuit is $0$ .

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