By connecting the ends of a battery with a resistance of 9 Omega, its potential difference reduces from 40 volts to 30 volts.

When the resistance of 9 Omega is connected at the ends of a battery, its potential difference decreases from 40 volt to 30 volt . The internal resistance of the battery is

The emf of a battery is 4.0 V and its internal resistance is 1.5 Omega . Its potential difference is measured by a voltmeter of resistance 1000 Omega . Calculate the percentage error in the reading of emf shown by voltmeter.

A resistance of 900Omega is connected in series with a galvanometer of resistance 100Omega . A potential difference of 1 Volt produces 100 division deflection in galvanometer. Find the figure of merit of galvanometer.

A capacitor of capacitance 1muF is connected in parallel with a resistance 10Omega and the combination is connected across the terminals of a battery of EMF 50 V and internal resistance 1Omega . The potential difference across the capacitor at steady state is (in Volt)

A resistance of 6 ohms is connected in series with another resistance of 4 ohms. A potential difference of 20 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 6 ohm resistance.

Two batteries of emf 4 V and 8 V with internal resistances 1Omega and 2Omega are connected in a circuit with a resistance of 9Omega as shown in figure. The current and potential difference between the points P and Q are

The charge stored in a capacitor of capacitance C, initially connected to a battery of V volts becomes half on reducing the potential difference across it by 100 V. Calculate the value of V.

Three resistances of magnitude 2, 3 and 5 ohm are connected in parallel to a battery of 10 volts and of negligible resistance. The potential difference across 3 Omega resistance will be