Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually.

Describe how a potentiometer is used to compare the e.m.f.s of two cells by connecting them separately.

Describe how potentiometer is used to compare the e.m.f.s of two cells by combination method.

Describe the use of potentiometer to compare the e.m.f. of two cells by the direct method (individual method).

Name the device used for measuring the e.m.f. of a cell.

When two cells of emfs. E_1 and E_2 are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.

A cell of e.m.f. 2 V and negligible internal resistance is connected in series with a potentiometer wire of length 100 cm. The e.m.f of the Leclanche cell is found to balance on 75 cm of the potentiometer wire. The e.m.f. of the cell is

A 2.0 V potentiometer is used to determine the internal resistance of 1.5 V cell. The balance point of the cell in the circuit is 75 cm . When a resistor of 10Omega is connected across cel, the balance point sifts to 60 cm . The internal resistance of the cell is

If two solar cells are connected in parallel then

Describe in detail the glial cells.

A potentiometer circuit has been setup for finding. The internal resistance of a given cell. The main battery used a negligible internal resistance. The potentiometer wire itsefl is 4 m long. When the resistance, R , connected across the given cell, has value of (i) Infinity 9.5 Omega , (ii) the 'balancing length' , on the potentiometer wire are found to be 3 m and 2.85 m , respectively. The value of internal resistance of the cell is