Terminal voltage of cell (emf = 3V & internal resistance = r) is equal to 2.5V and heat loss in R is given by 0.5Watt find power loss in internal resistance

A cell of emf 10 V and internal resistance 3 Omega is connected in parallel with another cell of emf 7V and internal resistance 3/5 Omega, such that their positive terminals are joined together and so there are negative terminals. Their positive terminals are joined with the negative terminal and their negative terminal is joined with the positive terminal of a third cell of emf 20 V with internal resistance 2Omega. The combination can be replaced by a battery of emf E and internal resistance r, then the values of E and R are respectively

A lead resistance R is connected across a cell of emf E and internal resistance r. If the closed circuit p.d. across the terminals of the cell is V the internal resistance of the cell is.

The emf of a cell is 6 V and internal resistance is 0.5 kOmega . What will be the reading (in volt) of a voltmeter having an internal resistance of 2.5kOmega ?

The emf of a cell is 6 V and internal resistance is 0.5 kOmega . What will be the reading (in volt) of a voltmeter having an internal resistance of 2.5kOmega ?

The emf of a cell is 6 V and internal resistance is 0.5 kOmega The reading of a Voltmeter having an internal resistance of 2.5 kOmega is

The emf of a cell is 6 V and internal resistance is 0.5 kOmega The reading of a Voltmeter having an internal resistance of 2.5 kOmega is

A cell of emf epsilon and internal resistance r gives a current of 0.5 A with an external resistance of 12 Omega and a current of 0.25 A with an external resistance of 25 Omega . Calculate (a) internal resistance of the cell and (b) emf of the cell.