In Fig, `E =5` volt , `r = 1 Omega, R_(2) = 4 Omega, R_(1) = R_(3) = 1 Omega` and `C = 3 muF`. Then the numbercal value of the charge on each plate of the capacitor is

In the circuit shown , the batteries have emf E_1 = E_2= 1V , E_3 = 2.5 V, and the resistance R_1 = 10Omega, R_2 = 20 Omega , Capacitance C = 10 muF . The charge on the left plate of the charge on the left plate of the capacitor C at steady state is

In the circuit shown in fig. R_(1) = 10 Omega, R_(2) = Omega, C_(1) = 1 muF, C_(2) = 2muF, and E =6V. Calculate the charge on each capactor in the stedy state.

In the circuit shown in figure, R(1) = 1Omega, R_(2) = 2Omega, C_(1) = 1 muF, C_(2) = 2 muF and E = 6 V . Calculate charge on each capacitor in steady state

In the circuit given E = 6.0 volt, R_(1) = 100Omega, R_(2) = R_(3)=50Omega and R_(4)=75Omega . The equivalent resistance of the circuit, in ohms is

In the circuit R_(1) = 4 Omega , R_(2) = R_(3) = 15 Omega R_(4) = 30 Omega and E = 10 V Calculate the equivalent resistance of the circuit and current in each resistor

Figure shows a battery with emf 15 V in a circuit with R_1 = 30Omega, R_2 = 10 Omega, R_3 = 20 Omega and capacitance C = 10 muF . The switch S is initially in the open position and is then closed at time t= 0. What will be the fimal steady - state charge on capacitor?

In the circuit shows in Fig E = 15 V , R_(1) = 1Omega , R_(2) = 1 Omega , R_(3) = 2Omega , and L = 1.5 H . The currents flowing through R_(1),R_(2) ,and R_(3) are i_(1),i_(2) , and i_(3) , respectively. After the circuit reaches the steady state,