In an AC circuit, the potential difference $V$ and current $I$ are given respectively by
$V = 100 sin (100 t) V$
and $I = 100 sin (100 t + \frac{π}{3}) m A$
The power dissipated in the circuit will be

10^{4} K W

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10 K W

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2.5 K W

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5 K W

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Solution

The correct option is C $2.5 K W$
$P o w e r = V π m s \times I_{R M S} \times cos θ$
$= \frac{V_{0} I}{2} \times cos θ$
$= \frac{100 \times 100}{2} \times \frac{1}{2}$
$= 2500 W$
$2.5 k W$ .

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In an AC circuit, the potential difference V and current I are given respectively byV=100sin(100t)Vand I=100sin(100t+π3)mAThe power dissipated in the circuit will be

The correct option is C 2.5KWPower=Vπms×IRMS×cosθ=V0I2×cosθ=100×1002×12=2500W2.5 kW.

In an AC circuit, the potential difference $V$ and current $I$ are given respectively by
$V = 100 sin (100 t) V$
and $I = 100 sin (100 t + \frac{π}{3}) m A$
The power dissipated in the circuit will be

The correct option is C $2.5 K W$
$P o w e r = V π m s \times I_{R M S} \times cos θ$
$= \frac{V_{0} I}{2} \times cos θ$
$= \frac{100 \times 100}{2} \times \frac{1}{2}$
$= 2500 W$
$2.5 k W$ .