Question

A cell is connected in series with a resistor and supplies a current of 4.0 A for a time of 500 s. During this time, 1.5 kJ of energy is dissipated in the cell and 2.5 kJ of energy is dissipated in the resistor. What is the emf of the cell?

The answer is 2 V.

Answer 1

emf = 2 V

Explanation:

te total energy dissipated is 1,5 kJ + 2,5 kJ = 4kJ
te energy that the cell supplies is the same :
`E= V xx I xx t= V xx 4.0 A xx 500 s = 4 kJ`
since`1 J = W xx s = 1 V xx 1 A xx 1 s`
hence you have
`V = E/(I xx t) = (4000J)/(2000 A xx s) = 2 V`