When an inductor L and a resistor R in series | vedclass

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(C) Given for the $LR$ series $AC$ circuit:Voltage, $V_{rms} = 12 \, V$Frequency, $f = 50 \, Hz$Current, $I = 0.5 \, A$Phase difference, $\phi = \frac

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$12$

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(C) Given for the $LR$ series $AC$ circuit:Voltage, $V_{rms} = 12 \, V$Frequency, $f = 50 \, Hz$Current, $I = 0.5 \, A$Phase difference, $\phi = \frac{\pi}{3}$First, calculate the total impedance $Z$ of the circuit using Ohm's law for $AC$ circuits:$Z = \frac{V_{rms}}{I} = \frac{12}{0.5} = 24 \, \Omega$The power factor of an $LR$ series circuit is given by:$\cos \phi = \frac{R}{Z}$Substituting the known values:$\cos(\frac{\pi}{3}) = \frac{R}{24}$Since $\cos(\frac{\pi}{3}) = 0.5$:$0.5 = \frac{R}{24}$$R = 24 	imes 0.5 = 12 \, \Omega$Therefore, the value of $R$ is $12 \, \Omega$.

When an inductor $L$ and a resistor $R$ in series are connected across a $12 \, V, 50 \, Hz$ supply, a current of $0.5 \, A$ flows in the circuit. The current differs in phase from the applied voltage by $\frac{\pi}{3}$ radian. Then the value of $R$ is (in $\Omega$)

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