Answer –
According to the question statement –
The EMF of storage battery is E = 8.0 V
Internal resistance of battery is given by r = 0.5 Ω
DC supply voltage is V = 120 V
Resistance of the resistor is R = 15.5 Ω
Effective voltage of the circuit = V 1
Since it is known that R is connected to the storage battery in series, we can write –
V 1 = V – E= 120 – 8
V 1 = 112 V
Current flowing in the circuit = I , it is given by the relation –
![\[I=\frac{{{V}^{1}}}{R+r}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-541fd4e6cd2458f6514d632e60614ff8_l3.png "Rendered by QuickLaTeX.com")
![\[I=\frac{112}{15.5+5}=\frac{112}{16}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2abd68314e63c9b209671bf855b531c8_l3.png "Rendered by QuickLaTeX.com")
Therefore, I = 7A
According to Ohm’s law, voltage can be given by the product –
I x R = 7 × 15.5
V = 108.5 V
We know that, DC supply voltage = Terminal voltage + voltage drop across R
Therefore, Terminal voltage of battery = 120 – 108.5 = 11.5 V
When used in a charging circuit, a series resistor restricts the amount of current drawn from the external source. In its absence, the current will become tremendously high. This is a really dangerous situation.