In the given circuit an ideal voltmeter reads 3 V and current through `2.5Omega` resistance is 2 A. EMF E of cell is:

In the given circuit, an ideal voltmeter connected across the 10Omega resistance reads 2V. The internal resistance r, of each cell is:

In the given circuit, the voltmeter records 5 V. The resistance of the voltmeter in Omega is

In the given circuit, the reading of ideal voltmeter is E//2 . The internal resistance of the battery is

In the given circuit, the reading of ideal voltmeter is E//2 . The internal resistance of the battery is

In the given circuit , the voltmeter reading is 4.5 V. Assuming that the voltmeter is ideal , current thorough 12Omega resistance is -

In the given circuit, the potential difference across the capacitor is 12V . Each resistance is of 3Omega . The cell is ideal. The emf of the cell is

Two cells each of emf 10 V and each 1Omega internal resistance are used to send a current through a wire of 2Omega resistance. The cells are arranged in parallel. Then the current through the circuit

Write any two factors on which internal resistance of a cell depends . The reading on a high resistance voltmeter , when a cell is connected across it , is 2.0 V . When the terminals of the cell are also connected to a resistance of 3 Omega as shown in the circuit , the voltmeter reading drops to 1.5 V . Find the internal resistance of the cell.

Write any two factors on which internal resistance of a cell depends . The reading on a high resistance voltmeter , when a cell is connected across it , is 2.0 V . When the terminals of the cell are also connected to a resistance of 3 Omega as shown in the circuit , the voltmeter reading drops to 1.5 V . Find the internal resistance of the cell.