A battery of emf 10 V and internal resistance 3`Omega` is connected to a resistor. If the current in the circuit is 0.5 A, find (i) the resistance of the resistor, (ii) the terminal voltage of the battery.

Explanation

Here emf `epsi`= 10 V, internal resistance r = 3`Omega` and current drawn I = .5 A
(i) As `I = (epsi)/(R + r) ` hence `0.5 = (10)/(R+3) rArr R = 17 Omega`
(ii) The internal voltage of battery `V = epsi - Ir = 10 - 0.5 xx 3 = 8.5 V`