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

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 potentiometer has uniform potential gradient across it. Two cells connected in series (i) to support each other and (ii) to oppose each other are balanced over 6 m and 2 m respectively on the potentiometer wire. The e.m.f.’s of the cells are in the ratio of

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

An electron in potentiometer experiences a force 2.4xx10^(-19)N . The length of potentiometer wire is 6m. The emf of the battery connected across the wire is (electronic charge =1.6xx10^(-19)C )

Derive the expression for E_eq and r_eq when two cells are connected in parallel.

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