As shown in the figure,a series circuit conne | vedclass

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(C) The total resistance of the circuit is $R = R_1 + R_2 = 44\,\Omega + 36\,\Omega = 80\,\Omega$.The net reactance of the circuit is $X = X_L - X_C =

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(C) The total resistance of the circuit is $R = R_1 + R_2 = 44\,\Omega + 36\,\Omega = 80\,\Omega$.The net reactance of the circuit is $X = X_L - X_C = 90\,\Omega - 30\,\Omega = 60\,\Omega$.The impedance of the circuit is $Z = \sqrt{R^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$.The current in the series circuit is $I = \frac{V}{Z} = \frac{200\, V}{100\,\Omega} = 2\, A$.The power dissipated in the coil is due to its internal resistance $R_2 = 36\,\Omega$. Therefore,$P = I^2 R_2 = (2)^2 	imes 36 = 4 	imes 36 = 144\, W$.

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