When a resistance of `2 Omega` is connected across the terminals of a cell, the current is 0.5 A. When resistance is increased to `5 Omega`, the current is 0.25 A. the emf of the cell is

When a resistance of 2 ohm is connected across terminals of a cell, the current is 0.5 A . When the resistance is increased to 5 ohm, the current is 0.25 A The e.m.f. of the cell is

When a resistance of 2 ohm is connected across terminals of a cell, the current is 0.5 A . When the resistance is increased to 5 ohm, the current is 0.25 A The e.m.f. of the cell is

When a resistance of 2 ohm is connected across terminals of a cell, the current is 0.5 A . When the resistance is increased to 5 ohm, the current is 0.25 A The e.m.f. of the cell is

When a resistance of 2 Ohm is connected across the terminals of a cell, the current is 0.5 A. If a resistance of 5 Ohm is connected across the cell then the current is 0.25 At the emf of the cell is:

When a resistance of 2 ohm is placed across a battery the current is 1 A and when the resistance across the terminals is 17 ohm, the current is 0.25A. The emf of the battery is

When a resistor of 20 Omega is connected in series with a battery, the current is 0.5 A. when a resistor of 10 Omega is connected the current becomes 0.8 A. calculate the emf and the internal resistance of the battery.

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

When a resistance of 12 Omega is connected with a cell of emf 1.5 V, 0.1A current flows through the resistance. Internal resistance of the cell is—