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

Explanation

Emf of the storage battery, `E = 8.0 V`
Internal resistance of the battery, `r - 0.5 Omega`
DC supply voltage, `V = 120 V`
Resistance of the resistor, `R = 15.5 Omega`
Effective voltage in the circuit `= V^(1)`
R is connected to the storage battery in series. Hence, it can be written as
`V^(1)=V-E`
`V^(1) = 120 - 8 = 112 V`
Current flowing in the circuit `= I`, which is given by the relation,
`I = (V^(1))/(R+r)`
`=(112)/(15.5 + 5)=(112)/(16)=7A`
Voltage across resistor R given by the product, `IR - 7 xx 15.5 = 108.5 V`
DC supply voltage = Terminal voltage of battery + Voltage drop acorss R
Terminal voltage of battery `= 120 - 108.5 = 11.5 V`
A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.