A storage battery of emf 8.0 V and internal resistance 0.5 `Omega` is being charged by a 120 V de supply using a series resistor of 15.5 `Omega` . What is the terminal voltage of the battery during charging ?

Explanation

Charging current is given by formula,
`I = (V - epsilon)/(R + r)`
= `(120 - 8)/(15.5 + 0.5) `
= ` (112)/(16) `
`therefore ` I = 7A
Terminal voltage of battery during its charging
`V. = epsilon` + Ir
= 8 + (7) (0.5)
8 + 3.5
`therefore V. = 11.5` V
By connecting series resistance, we can reduc charging current (with the increase in chargin, time). By reducing charging current, we reduce the heat loss in the internal resistance.