An a.c. source of 15 , V, 50 , Hz is connecte | vedclass

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(C) Given data: $E = 15 \, V$, $f = 50 \, Hz$, $I = 0.5 \, A$, $\phi = \frac{\pi}{3} \, rad$.The impedance of the circuit is given by $Z = \frac{E}{I}

Vedclass · Accepted Answer

$15$

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(C) Given data: $E = 15 \, V$, $f = 50 \, Hz$, $I = 0.5 \, A$, $\phi = \frac{\pi}{3} \, rad$.The impedance of the circuit is given by $Z = \frac{E}{I} = \frac{15}{0.5} = 30 \, \Omega$.The phase difference $\phi$ in an $LR$ circuit is given by $	an \phi = \frac{X_L}{R}$.Substituting the values: $	an \frac{\pi}{3} = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = \sqrt{3}R$.The impedance formula is $Z = \sqrt{R^2 + X_L^2}$.Substituting $X_L = \sqrt{3}R$: $Z = \sqrt{R^2 + (\sqrt{3}R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.Since $Z = 30 \, \Omega$, we have $2R = 30 \, \Omega$.Therefore, $R = \frac{30}{2} = 15 \, \Omega$.

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