Question

a battery of internal resistance 0.5 ohm is connected in a circuit of three resistances,as shown. A current of 2A is flowing in the circuit. Calculate:(i) emf of the battery, (ii) potential difference across the ends of the 5 ohm resistance​

Answer 1

Answer:

Vt = 10.5 v & R = 21 ohm

Explanation:

E = 12 v

r = 3 ohm

I = 0.5 A

Vt = E - Ir

Vt = 12 - 0.5x3

Vt = 10.5 v

Vt = IR

R = Vt / I

R = 10.5 / 0.5

R = 21 ohm

Answer 2

Answer

id k then

Explanation: