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(C) Given: Resistance $R = 100 \, \Omega$,Capacitive reactance $X_C = 100 \, \Omega$,and $RMS$ voltage $V_{rms} = 220 \, V$.The total impedance of the

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$2.2$

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(C) Given: Resistance $R = 100 \, \Omega$,Capacitive reactance $X_C = 100 \, \Omega$,and $RMS$ voltage $V_{rms} = 220 \, V$.The total impedance of the $RC$ series circuit is $Z = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \, \Omega$.The peak voltage of the source is $V_0 = V_{rms} \sqrt{2} = 220\sqrt{2} \, V$.The peak conduction current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{100\sqrt{2}} = 2.2 \, A$.According to Maxwell's modification of Ampere's law,the displacement current $I_d$ between the plates of a capacitor is equal to the conduction current $I_c$ in the connecting wires.Therefore,the peak value of the displacement current is equal to the peak value of the conduction current,which is $2.2 \, A$.

$A$ $100 \, \Omega$ resistance and a capacitor of $100 \, \Omega$ reactance are connected in series across a $220 \, V$ source. When the capacitor is $50\%$ charged,the peak value of the displacement current is.....$A$

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