A storage battery of emf 8.0 V and internal resistance `0.5 Omega` is being charged by a 120 V dc 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

Given that `E = 8.0V, V =120V, r = 0.5Omega, and R =15.5Omega`
Current in the circuit during charging is given by
`I=("Total voltage")/("Total resistance") =(V- E)/(R -r) = (120 - 8)/(15.5 + 05) = (112)/(16) = 7Omega`
During charging , the current flows inside the battery in a direction
opposite to that during discharge. Hence, the terminal voltage
of the battery during charging is
`E +Ir = 8 +7xx0.5 = 11.5V`
A series resistor in the charging circuit limits the current drawn from the external source. in its absence, the current will be dangerously high.