A circuit when connected to an AC source of 12 ~V gives a current...

When the circuit is connected to AC source, Voltage, $\mathrm{V}=12 \mathrm{~V}$ Current, $\mathrm{I}=0.2 \mathrm{~A}$ $\Rightarrow$ Impedance \mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.2}=60 \Omega When it is connected to DC source, Voltage, $\mathrm{V}=12 \mathrm{~V}$ Current, $\mathrm{I}=0.4 \mathrm{~A}$ $\Rightarrow$ Resistance, \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.4}=30 \Omega As in case of DC supply, the capacitor act as an open circuit and no current flows through the circuit. So the given circuit will not have capacitor in series combination. Therefore the circuit should be a series LR circuit.