Question

A storage battery of emf 8.0 V and internal resistance 0.5 W is beingcharged by a 120 V dc supply using a series resistor of 15.5 W. Whatis the terminal voltage of the battery during charging? What is thepurpose of having a series resistor in the charging circuit?

Answer 1

Given: The emf of the storage battery is 8.0 V, the internal resistance of the battery is 0.5 Ω the supply voltage is 120 V, the resistance of the series resistor is 15.5 Ω.

The effective voltage in the circuit is given as,

V ′ =V−E

Where, the emf of the storage battery is E, the supply voltage is V and the effective voltage in the circuit is V ′ .

By substituting the given values in the above equation, we get

V ′ =120−8 =112 V

The current flowing in the circuit is given as,

I= V ′ R+r

Where, the internal resistance is r, the resistance of the series resistor is R and the current flowing in the circuit is I.

By substituting the given values in the above formula, we get

I= 112 15.5+.5 = 112 16 =7 A

The voltage across the resistor is given as,

V ″ =IR

By substituting the given values in the above formula, we get

V ″ =IR =7×15.5 =108.5 V

The terminal voltage is given as,

V t =V− V ″

By substituting the given values in the above formula, we get

V t =120−108.5 V t =11.5 V

Thus, the terminal voltage of the battery is 11.5 V.

In a charging circuit the purpose of the series resistor is that it limits the current drawn from the external source. In the absence of series resistor, the current will be extremely high and the high current is very dangerous for the circuit.