In an AC circuit V and I are given by V-100sin100tvolt- I-100sin100t-3amp the power dissipated in the circuit is

The correct option is B 2.5 kW P=V_rms×I_rms×cosϕ =V_∘I_∘2cosϕ =100× 1002×12 =2500 W =2.5 kW ...

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Standard IX

Physics

Direct Current vs. Alternating Current

In an AC circ...

Question

In an AC circuit $V$ and $I$ are given by $V = 100 sin (100 t)$ volt, $I = 100 sin (100 t + \frac{π}{3})$ amp the power dissipated in the circuit is

5.0 k W

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2.5 k W

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1.25 k W

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z e r o

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Solution

The correct option is B $2.5 k W$
$P = V_{r m s} \times I_{r m s} \times cos ϕ$
$= \frac{V_{\circ} I_{\circ}}{2} cos ϕ$
$= \frac{100 \times 100}{2} \times \frac{1}{2}$
$= 2500$ W
$= 2.5$ kW

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In an AC circuit V and I are given by V=100sin(100t)volt, I=100sin(100t+π3)amp the power dissipated in the circuit is

The correct option is B 2.5 kWP=Vrms×Irms×cosϕ=V∘I∘2cosϕ=100×1002×12=2500W=2.5kW

In an AC circuit $V$ and $I$ are given by $V = 100 sin (100 t)$ volt, $I = 100 sin (100 t + \frac{π}{3})$ amp the power dissipated in the circuit is

The correct option is B $2.5 k W$
$P = V_{r m s} \times I_{r m s} \times cos ϕ$
$= \frac{V_{\circ} I_{\circ}}{2} cos ϕ$
$= \frac{100 \times 100}{2} \times \frac{1}{2}$
$= 2500$ W
$= 2.5$ kW