The current and voltage functions in an AC circuit are i - 100sin 100tmA - V - 100sin 100t - -3V The power dissipated in the circuit is

The correct option is B 2.5 W i=100sin 100tmA v=100 sin(100t+x/3)V P=I_rms× V_rms cosθ =100/√(2)×100/√(2)×1/2× 10^ 3W =10/4=2· 5W ...

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Standard XII

Physics

Power in an AC Circuit

The current a...

Question

The current and voltage functions in an AC circuit are $i = 100 sin 100 t m A$ , $V = 100 sin (100 t + \frac{π}{3}) V$ The power dissipated in the circuit is

10 W

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2.5 W

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5 W

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5 kW

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Solution

The correct option is B 2.5 W
$i = 100 s i n 100 t m A$
$v = 100 s i n (100 t + \frac{x}{3}) V$
$P = I_{r m s} \times V_{r m s} c o s θ$
$= \frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \frac{1}{2} \times 10^{- 3} W$
$= \frac{10}{4} = 2 \cdot 5 W$

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The current and voltage functions in an AC circuit arei=100sin100tmA , V=100sin(100t+π3)V The power dissipated in the circuit is

The correct option is B 2.5 Wi=100sin100tmAv=100sin(100t+x3)VP=Irms×Vrmscosθ=100√2×100√2×12×10−3W=104=2⋅5W

The current and voltage functions in an AC circuit are $i = 100 sin 100 t m A$ , $V = 100 sin (100 t + \frac{π}{3}) V$ The power dissipated in the circuit is

The correct option is B 2.5 W
$i = 100 s i n 100 t m A$
$v = 100 s i n (100 t + \frac{x}{3}) V$
$P = I_{r m s} \times V_{r m s} c o s θ$
$= \frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \frac{1}{2} \times 10^{- 3} W$
$= \frac{10}{4} = 2 \cdot 5 W$