In the diagram shown, reading of the ammeter just after closing the switch S is 12 A and a long time after S is closed is 4 A. Then value of ‘R’ is

Just after S is closed, charge on the capacitor is zero. So equivalent resistance =(2R2+R)Ω∴ E2R/(2+R)=12⇒(2+R)E=24R.....(1)A long time after S is closed, the capacitor becomes fully charged. There fore E2=4⇒E=8 volt ......(2)∴(2+R)8=24R⇒R=1Ω