In the circuit shown below the cell is ideal, with emf = 2 V: The resistance of the coil of the galvanometer G is `1 Omega`. Then, find the potential difference across `C_(1)` (in volts), in steady state

In the circuit shows in Fig. 6.63, the cell is ideal with emf 9 V . If the resistance of the coil of galvanometer is 1 Omega , then

In the circuit shown, the cell is ideal with emf = 2V . The resistance of the coil of the galvanometer G is 1 Omega . (i) No current flows in G (ii) 0.2A current flows in G (iii) potential difference across C_(1) is 1V (iv) Potential difference across C_(2) is 1.2V

In the circuit shown, the cell is ideal , with emf= 15 V . Ecah resistance is of 3 Omega. the potential difference across the capacitor is

In the circuit shown below, the cell has an e.m.f. of 10 V and internal resistance of 1 ohm . The other resistances are shown in the figure. The potential difference V_(A) - V_(B) is

In the circuit shown in fig. the cell has emf 10V and internal resistance 1 Omega

In the circuit shown in fig. the cell is ideal with emf 15V. Each resistance is of 3Omega. The potential difference across the capacitoro in steady state is

In the circuit shown below: The potential difference across the 3 Omega resistor is:

In the fig. shown the circuit is in steady state. what is the potential difference across the capacitor in steady state ?

In the given circuit diagram find the potential difference across 2 Omega resistance (in volts).

In the electric circuit shown each cell has an emf of 2 V and internal resistance of 1Omega . The external resistance is 2Omega . The value of the current is (in A)