In the given circuit switch K is open. The charge on the capacitor is C is steady-state is `q_1` Now the key is closed and steady-state charge on C is `q_2` The ratio of charges `q_1//q_2` is -

In the circuit shown in the figure K_(1) is open.The charge on capacitor C in steady state is q_(1) .Now key is closed and at steady state charge on C is q_(2) . The ratio of charges q_(1)/q_(2) is qquad R_(1)=2Omega E

In given circuit, switch S is closed at t=0 . The charge on the capacitor ar steady state will be

The charge on the capacitor in steady state in the circuit shown is

The charge on the capacitor in steady state in the circuit shown is

The circuit is in a steady state. The charge in capacitor C_1 is

The circuit is in a steady state. The charge in capacitor C_1 is

The circuit is in a steady state. The charge in capacitor C_2 is

The circuit is in a steady state. The charge in capacitor C_2 is