In the circuit, $ E_1 = E_2 = E_3 = 2 \, \text{V} $ and $ R_1 = R_2 = 4 \, \Omega $. Then the current flowing through $ E_2 $ is:
Switch $ S $ is closed. Now the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is:
The resultant resistance between $ A $ and $ B $ in the given figure is: