A Cell sends a current of 0.5 A through a `2 Omega` resistance. If the resistance is increased by `5 Omega`, the current falls to 0.25 A. Find the emf of the cell and its internal resistance.

Write relationship between e.m.f. of a cell and its internal resistance.

Find out the value of current through 2 Omega resistance for the given circuit

The emf of a battery is 21 volt. When it sends a current of 1 ampere through an external resistance, the P.D. across its terminalsfalls to 19 volt. What is the internal resistance of battery?

A cell supplies a current 1.2A through a resistor 2 Omega . The same cell supplies a current 0.4A through a 7 Omega resistor. The internal resistance of the cell is

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Omega the balance point is found at 40 cm of the wire from the same end. The internal resistance of the cell is

The current flowing through an external resistance of 2 Omega is 0.5 A when it is connected to the terminals of a cell. This current reduces to 0.25 A when the external resistance is 5 Omega .Use Kirchhoff's laws to find e.m.f. of cell

The combined resistance of a galvanometer of resistance 495 Omega and its shunt is 9.9 Omega . Find the shunt resistance.

In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across 0.52 m of the potentiometer wire. If the cell is shunted by a resistance of 5Omega balance is obtained when the cell connected across 0.4m of the wire. Find the internal resistance of the cell.

In a potentiometer experiment, e.m.f. of a cell is balanced by a length of 150 cm on the potentiometer wire. When the cell is shunted by a 10 Omega resistance, the balancing length reduces to 90 cm. What is the internal resistance of the cell?