EMF of an electrical cell is 2 volt. A `10 Omega` resistance is joined at its two ends then potential difference is measured 1.6 volt. Find out the internal resistance and lost volt.

The potential difference in open circuit for a cell is 2.2 volts. When a 4 ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will be

The potential difference in open circuit for a cell is 2.2 votls . When a 4 ohm resistor is connected between its two electrodes that potential difference becomes 2 volts. The internal resistance of cell will be

The e.m.f. of a cell is 6 volt. When 2 ampere current is drawn from it then th e potential difference across its terminal remains 3 volt. Its internal resistance.

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

Two resistance of 400 Omega and 800 Omega are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Omega is used to measure the potential difference across 400 Omega . The error in measurement of potential difference in volts approximatley is

In the electrical network shown in the figure, What will be the potential difference (in volt) across 3Omega resistance ?

A current of 5 amperes flows through a wire whose ends are at a potential difference of 3 volts. Calculate the resistance of the wire.