A galvanometer G of 2 Ω resistance is connected in the given circuit. The ratio of charge stored in C 1 and C 2 is:

Step-by-step solution

In steady state: R e q = 12 Ω Current through the branches with resistance, I = 6 12 = 0 . 5 A Now, potential drop across C 1 = 3 V and potential drop C 2 = 4 V Charge on the first capacitor, q 1 = C 1 V 1 = 12 μ C Charge on the second capacitor, q 2 = C 2 V 2 = 24 μ C Therefore, q 1 q 2 = 1 2