A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be

P (\frac{R}{Z})

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P

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P {(\frac{R}{Z})}^{2}

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P \sqrt{\frac{R}{Z}}

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Solution

The correct option is B $P {(\frac{R}{Z})}^{2}$
Power $P = i_{r m s}^{2} R = (\frac{V_{r m s}}{R})^{2} R$
Power when inductor is put in series with R: $P^{'} = (\frac{V_{r m s}}{Z})^{2} R$
$\frac{P^{'}}{P} = \frac{\frac{R}{Z^{2}}}{\frac{R}{R^{2}}} = (\frac{R}{Z})^{2}$
$P^{'} = P (\frac{R}{Z})^{2}$

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A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be

The correct option is B P(RZ)2Power P=i2rmsR=(VrmsR)2RPower when inductor is put in series with R: P′=(VrmsZ)2RP′P=RZ2RR2=(RZ)2P′=P(RZ)2

The correct option is B $P {(\frac{R}{Z})}^{2}$
Power $P = i_{r m s}^{2} R = (\frac{V_{r m s}}{R})^{2} R$
Power when inductor is put in series with R: $P^{'} = (\frac{V_{r m s}}{Z})^{2} R$
$\frac{P^{'}}{P} = \frac{\frac{R}{Z^{2}}}{\frac{R}{R^{2}}} = (\frac{R}{Z})^{2}$
$P^{'} = P (\frac{R}{Z})^{2}$