Question

A current of 2A is flowing through a cell of e.m.f 5V and internal resistance 0.5ohm from negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive electrode will be:-
(a)5V (b)14V
(c)15V
(d) 16V

Answer 1

Answer ⇒ The Emf of Positive Electrode is 15 V. Option (c).

Explanation ⇒ Since, current is flowing from negative to positive electrode, this means cell is in an state of charging.

Thus,

V = E + ir

∴ iR = E + ir

∴ R = (ir + E)/i

∴ R = (2 × 0.5 + 5)/2

∴ R = 6/2

∴ R = 3 ohm.

∴ R = 3 Ω.

Hence, the external resistance is 3 Ω.

Now, EMF OF CELL = Emf of Positive Electrode - Emf of negative electrode.

∴ 5 = Emf of Positive Electrode - 10

∴ Emf of Positive Electrode = 10 + 5

∴ Emf of Positive Electrode = 15 V.

Hence, the Emf of Positive Electrode is 15 V.

Hence, Option (c). is correct.

Hope it helps.