A storage battery of emf 0.8 V and internal resistance `0.5 Omega` is being charged by a 120 V do supply using a series resistor of `15.5 Omega`. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Explanation

Here emf of the battery = 0.8 V, voltage of d.c. supply = 120 V
Internal resistance of battery `r = 0.5 Omega`, external resistance `R = 15.5 Omega`
Since a storage battery of emf 8V is charged with a.d.c supply of 120 V effective emf in the circuit is given by `epsilon = 120 - 8 = 112V`
Total resistance of the circuit `= R + r = 15.5 + 0.5 = 16 Omega`
`:.` current in the circuit during charging is given by
`I = (epsilon)/(R + r) = (112)/(16) = 7.0 A`
`:.` Voltage across `R = IR = 7.0 xx 15.5 = 108.5 V`
During charging the voltage of the d.c supply in a circuit must be equal to the sum of the voltage drop across R and terminal voltage of the battery
`:.` 120 = 108.5 V or V = 120 - 108.5 = 11.5V
The series resistor limits the current drawn from the external source of d.c supply. In its absence the current will be dangerously high.