Internal Resistance Of A Cell

In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Omega resistance and is at a length of 3 m when the cell is shunted by a 10 Omega resistance, the internal resistance of the cell is then

In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shuntedd by a 5 Omega resistance, and is at a length of 3 m when the cell is shunted by a 10 Omega resistance. The internal resistance of the cell is, then

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.

In an experiment to find the internal resistance of a cell by a potentiometer, a balance was obtained for 50 cm length of the potentiometer wire, with a cell of e.m.f. 2V. When the cell was shunted by a resistance of 2 Omega , the balancing length of the potentiometer wire was 40 cm. What was the internal resistance of the cell ?

In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 4Omega resistance and at 3 m when cell is shunted by a 8Omega resistance. The internal resistance of cell is -

The internal resistance of the cell can be determined by,

A cell develops the same power across two resistances R_(1) and R_(2) separately. The internal resistance of the cell is

A cell of constant emf first connected to a resistance R_(1) and then connected to resistance R_(2) . If power deliverd in both cases is same, then the internal resistance of the cell is