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

Resistance of 60 watt bulb connected with 220 volt potential difference

The potential difference between the terminals of a cell, in an open circuit was 2.5 volt and when it was connected through a resistance of 5 ohm, the potential difference fell to 1.5v. Find the internal resistance of the cell.

Name the quantity for which the potential difference of a cell becomes less than its emf due to its internal resistance.

Emf of an electric cell is 1.5 V and its internal resistance Is 2 Omega . With this cell 1 Omega, 2 Omega and 10 Omega resistances are arranged in series .Determine the terminal potential differences of the resistances and the lost volt.

In a potentiometer experiment , a particular cell produces balance in the wire at 560 cm . If a 10Omega resistance is connected with the cell in parallel ,then this balanced lendth change to 412 cm . Find out the internal resistance of the cell .

The electromotive force of a cell is 2V. The potential difference becomes 1.5 V when a resistance of 15 Omega is added to the two ends of the cell. Determine the internal resistance of the cell and the lost volt.

A cell of emf 1.4V and internal resistance 2 Omega is connected in series with a resistance of 100 Omega and an ammeter. The resistance of the ammeter is 4/3 Omega .To measure the potential difference between the two ends of the resistance a voltmeter is connected IF the reading of the voltmeter is 1.10V, what will be its error?

A cell of emf 1.4V and internal resistance 2 Omega is connected in series with a resistance of 100 Omega and an ammeter. The resistance of the ammeter is 4/3 Omega .To measure the potential difference between the two ends of the resistance a voltmeter is connected IF the reading of the ammter is 0.02 A.what is the resistance of the voltmeter?

A cell of emf 1.4V and internal resistance 2 Omega is connected in series with a resistance of 100 Omega and an ammeter. The resistance of the ammeter is 4/3 Omega .To measure the potential difference between the two ends of the resistance a voltmeter is connected .Draw the circuit