The potential difference across the terminals of a battery is 8.70 VV when there is a current of 1.47 AA in the battery from the negative to the positive terminal. When the current is 3.42 AA in the reverse direction, the potential difference becomes 10.50 VV.(a) What is the internal resistance of the battery?
(b) What is the emf of the battery?

Explanation:

in case of charging

V = 8.7 V, i = 1.47 A

in case of discharging

V = 10.5 V, i = 3.42 A

Let E be the emf of the cell and r be the internal resistance

So, E = 8.7 + 1.47 r (1)

E = 10.5 - 3.42 r (2)

By equation (1) and (2)

8.7 + 1.47 r = 10.5 - 3.42 r

4.89 r = 1.8

r = 0.36 ohm

Put in equation (1)

E = 8.7 + 1.47 x 0.36

E = 9.23 V

2b2t

explanation:

2b2t

no sign of dia of fuel cell