A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The circuit is

When the circuit is connected to an AC
source of 12 V,
Current ${\mathrm{I}}_{\mathrm{rms}}=0.2\mathrm{A} \mathrm{for} {\mathrm{E}}_{\mathrm{rms}}=12\mathrm{V}.$
We know, ${\mathrm{E}}_{\mathrm{rms}}={\mathrm{I}}_{\mathrm{rms}}\times \mathrm{Z}$
Where Z is impedence
$\Rightarrow \mathrm{Z}=\frac{{\mathrm{E}}_{\mathrm{rms}}}{{\mathrm{I}}_{\mathrm{rms}}}=\frac{12}{0.2}=60\mathrm{\Omega }$
When the same circuit is connected to DC source, current produced, I = 0.4 A
There is current in steady state, which is not possible if there is a capacitor.
 We know, V = IR Where V is DC source voltage
I is current
R is resistance
$\Rightarrow \mathrm{R}=\frac{\mathrm{V}}{\mathrm{l}}=\frac{12}{0.4}=30 \mathrm{\Omega }$
$\therefore \mathrm{Z}>\mathrm{R};$there is another element in the circuit, increasing the impedence. It can’t be a capacitor, it should be an inductor.
$\therefore$The circuit is an L–R circuit.