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Byju's Answer

Standard XII

Physics

Power in an AC Circuit

An inductor o...

Question

An inductor of inductance $L$ and resistor of resistance $R$ are joined in series and connected by a source of frequency $ω$ . Power dissipated in the circuit is:

\frac{(R^{2} + ω^{2} L^{2})}{V}

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\frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}

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\frac{V}{(R^{2} + ω^{2} L^{2})}

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\frac{\sqrt{R^{2} + ω^{2} L^{2}}}{V^{2}}

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Solution

The correct option is B $\frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}$
As we know,

$P = V i c o s ϕ$
$= V (\frac{V}{Z}) (\frac{R}{Z})$
$= \frac{V^{2} R}{Z^{2}} = \frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}$

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An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency ω. Power dissipated in the circuit is:

The correct option is B V2R(R2+ω2L2)As we know,P=Vicosϕ=V(VZ)(RZ)=V2RZ2=V2R(R2+ω2L2)

An inductor of inductance $L$ and resistor of resistance $R$ are joined in series and connected by a source of frequency $ω$ . Power dissipated in the circuit is:

The correct option is B $\frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}$
As we know,

$P = V i c o s ϕ$
$= V (\frac{V}{Z}) (\frac{R}{Z})$
$= \frac{V^{2} R}{Z^{2}} = \frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}$