An inductor of reactance 100 Omega, a capaci | vedclass

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(B) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Given $R = 50 \ \Omega$, $X_L = 100 \ \Omega$, and $X_C

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$1$

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(B) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Given $R = 50 \ \Omega$, $X_L = 100 \ \Omega$, and $X_C = 50 \ \Omega$.$Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \ \Omega$.The average power dissipated in an $AC$ circuit is $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi = \frac{R}{Z}$ is the power factor.$P = V_{rms} 	imes \left( \frac{V_{rms}}{Z} ight) 	imes \left( \frac{R}{Z} ight) = \frac{V_{rms}^2 R}{Z^2}$.Given $V_{rms} = 10 \ V$.$P = \frac{10^2 	imes 50}{(50\sqrt{2})^2} = \frac{100 	imes 50}{2500 	imes 2} = \frac{5000}{5000} = 1 \ W$.

An inductor of reactance $100 \ \Omega$, a capacitor of reactance $50 \ \Omega$, and a resistor of resistance $50 \ \Omega$ are connected in series with an $AC$ source of $10 \ V, 50 \ Hz$. Average power dissipated by the circuit is . . . . . . $W$.

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