Question

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

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Solution

Power of an alternating circuitPower of an alternating circuit carrying current I with resistance R, impedance V is defined by the form, $P={\left(\frac{{V}_{rms}}{Z}\right)}^{2}R$, where, ${V}_{rms}$ is the square the root of the time average of the voltage squared.Impedance of an LR circuit is $Z=\sqrt{\left({R}^{2}+{\left(\omega L\right)}^{2}\right)}$, where, $R$ is the resistance and $\omega L$ is the inductive reactance due to the inductor L.DiagramPower of the circuit In the absence of the inductance, the power of the circuit is given by$P={\left(\frac{{V}_{rms}}{R}\right)}^{2}R................\left(1\right)$.When the inductance is connected with the resistance in series, then power is $P\text{'}={\left(\frac{{V}_{rms}}{Z}\right)}^{2}R................\left(2\right)$Comparing (1) and (2) we get,$\frac{P\text{'}}{P}=\frac{{\left(\frac{{V}_{rms}}{Z}\right)}^{2}R}{{\left(\frac{{V}_{rms}}{R}\right)}^{2}R}={\left(\frac{R}{Z}\right)}^{2}\phantom{\rule{0ex}{0ex}}orP\text{'}=P{\left(\frac{R}{Z}\right)}^{2}$Therefore, the power will be $P\text{'}=P{\left(\frac{R}{Z}\right)}^{2}$.

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