In an ac circuit- V and I are given by V-50sin 50t and I-100sin 50t- -3 mA- The power dissipated in the circuit-

The correct option is B 1.25W Dissipated power, P=V_rms× I_rmscosϕ Here, V_rms=V_0√(2)=50√(2) ; I_rms=I_0√(2)=100√(2) and cosϕ=cosπ/3= ...

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Standard XII

Physics

Electric Generator

In an ac circ...

Question

In an ac circuit, $V$ and $I$ are given by $V = 50 sin 50 t$ and $I = 100 sin (50 t + π / 3) m A$ . The power dissipated in the circuit.

2.5 W

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1.25 W

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5.0 W

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500

watt

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Solution

The correct option is B $1.25 W$
Dissipated power, $P = V_{r m s} \times I_{r m s} cos ϕ$
Here, $V_{r m s} = \frac{V_{0}}{\sqrt{2}} = \frac{50}{\sqrt{2}}$ ; $I_{r m s} = \frac{I_{0}}{\sqrt{2}} = \frac{100}{\sqrt{2}}$ and $cos ϕ = cos π / 3 = 1 / 2$
Thus, $P = \frac{50}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \frac{1}{2} = 1250 m W = 1.25 W$ (as unit of V is volt and unit of current mA).

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In an ac circuit, V and I are given by V=50sin50t and I=100sin(50t+π/3)mA. The power dissipated in the circuit.

The correct option is B 1.25WDissipated power, P=Vrms×IrmscosϕHere, Vrms=V0√2=50√2; Irms=I0√2=100√2 and cosϕ=cosπ/3=1/2Thus, P=50√2×100√2×12=1250mW=1.25W (as unit of V is volt and unit of current mA).

In an ac circuit, $V$ and $I$ are given by $V = 50 sin 50 t$ and $I = 100 sin (50 t + π / 3) m A$ . The power dissipated in the circuit.