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(C) In the $AC$ circuit, the current is in phase with the emf, which implies that the circuit is in resonance. At resonance, the impedance $Z$ of the

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$12$

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(C) In the $AC$ circuit, the current is in phase with the emf, which implies that the circuit is in resonance. At resonance, the impedance $Z$ of the circuit is equal to the resistance $R$ of the inductor coil $(Z = R)$.Given, rms voltage $V_{	ext{rms}} = 8 \, V$ and rms current $I_{	ext{rms}} = 16 \, A$.The resistance of the inductor coil is $R = \frac{V_{	ext{rms}}}{I_{	ext{rms}}} = \frac{8}{16} = 0.5 \, \Omega$.When the inductor coil is connected to a $6 \, V$ $DC$ battery, the capacitor acts as an open circuit for $DC$, but the question asks for the steady current through the inductor coil itself. Since the inductor coil has a resistance $R = 0.5 \, \Omega$, the steady current $I$ is given by Ohm's law:$I = \frac{V_{	ext{DC}}}{R} = \frac{6 \, V}{0.5 \, \Omega} = 12 \, A$.

An inductor coil is connected to a capacitor and an $AC$ source of rms voltage $8 \, V$ in series. The rms current in the circuit is $16 \, A$ and is in phase with the emf. If this inductor coil is connected to a $6 \, V$ $DC$ battery, the magnitude of the steady current is (in $A$)

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