An inductor of reactance 1 Omega and a resist | vedclass

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(B) The average power dissipated in an $LR$ series circuit is given by the formula:$P_{avg} = I_{rms}^2 R = \frac{V_{rms}^2 R}{Z^2}$Here,$Z$ is the im

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$30$

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(B) The average power dissipated in an $LR$ series circuit is given by the formula:$P_{avg} = I_{rms}^2 R = \frac{V_{rms}^2 R}{Z^2}$Here,$Z$ is the impedance of the circuit,given by $Z = \sqrt{X_L^2 + R^2}$.Given values are $X_L = 1 \ \Omega$,$R = 3 \ \Omega$,and $V_{rms} = 10 \ V$.First,calculate the impedance $Z$:$Z = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \ \Omega$.Now,calculate the average power $P_{avg}$:$P_{avg} = \frac{(10)^2 	imes 3}{(\sqrt{10})^2} = \frac{100 	imes 3}{10} = 30 \ W$.Thus,the power dissipated in the circuit is $30 \ W$.

An inductor of reactance $1 \Omega$ and a resistor of resistance $3 \Omega$ are connected in series to the terminals of a $10 \ V$ (rms) $AC$ source. The power dissipated in the circuit is (in $W$)

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