Statement I: Internal resistance of a battery is drawn parallel to a battery in electical circuit.
Statement II: Heat generated in a battery is due to internal resistance.

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-

In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire then internal resistance r of the battery is

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-

cells each of emf e and internal resistance r are joined in series to form a row and such rows are connected in parallel to form a battery of N = mn cells. If this battery is connected to an external resistance R . What is the emf of this battery and how much is its internal resistance ?

The current in a circuit containing a battery connected to 2Omega resistance is 0.9A. When a resistance of 7Omega is connected to the same battery, the current observed in the circuit is 0.3A. Then the internal resistance of the battery.

The current in a circuit containing a battery connected to 2Omega resistance is 0.9A. When a resistance of 7Omega is connected to the same battery, the current observed in the circuit is 0.3A. Then the internal resistance of the battery.

For maximum power from battery the internal resistance of battery r is