When the modified galvanometer of previous question is connected across te terminal of a battery, it shows a current of 4A. The current drops to 1A When resistance of `1.5Omega` is connected in series withh the modified galvanometer. Find the emf and the internal resistance of the battery.

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 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.

When a battery is connected to the resistance of 10Omega the current in the circuit is 0.12 A. The same battery gives 0.07A current with 20Omega Calculate e.m.f. and internal resistance of the battery.

When galvanometer of unknown resistance connected across a series combination of two identical batteries each of 1.5 V, the current through the resistor is 1A. When it is connected across parallel combination of the same batteries, the current through it is 0.6 A. The internal resistance of each battery is