Question

A storage battery of emf 8.0V and internal resistance 0.5 $ \Omega $ is being charged by a 120V 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?

Answer 1

Hint: Knowledge of basic electronics is required to solve this problem. We must also know, how to find the effective current in the circuit in this case, given by $ {{I}_{eff}}=\dfrac{{{V}_{eff}}}{{{R}_{i}}+{{R}_{series}}} $. Knowing that the terminal voltage of the battery, means that the voltage drop in between the terminals of the battery $(emf)$ is important, which includes the internal resistance $ ({{V}_{i}})$ in it as well, ${{V}_{ter\min al}}=emf+{{V}_{i}}$.

Complete step-by-step answer:
Given, the emf of the battery is, $ emf=8.0V $ and internal resistance of the battery is, $ {{R}_{i}}=0.5\Omega . $
The external dc supply is, $ {{V}_{ext}}=120V $ which is charging the battery.
A resistor is connected in between the dc charging supply and the battery, ${{R}_{series}}=15.5\Omega .$
Hence, the effective potential is, ${{V}_{eff}}={{V}_{ext}}-emf\Rightarrow {{V}_{eff}}=120-8\Rightarrow {{V}_{eff}}=112V. $
This means, that the effective current in the circuit is, $ {{I}_{eff}}=\dfrac{{{V}_{eff}}}{{{R}_{i}}+{{R}_{series}}}\Rightarrow {{I}_{eff}}=\dfrac{112}{0.5+15.5}\Rightarrow {{I}_{eff}}=\dfrac{112}{16}\Rightarrow {{I}_{eff}}=7A. $
Hence, the voltage drop across the internal resistance is\[{{V}_{i}}={{I}_{eff}}\times {{R}_{i}}\Rightarrow {{V}_{i}}=7\times 0.5\Rightarrow {{V}_{i}}=3.5V.\]
Therefore, the terminal voltage of the battery during the charging is $ {{V}_{ter\min al}}=emf+{{V}_{i}}\Rightarrow {{V}_{ter\min al}}=8+3.5\Rightarrow {{V}_{ter\min al}}=11.5V. $
 Coming to the next part of the question, the series resistor in a charging circuit is necessary to limit the amount of current drawn from the external dc source. If the series resistor wasn’t available, then the current flowing in the circuit would have been $ {{I}_{eff}}=\dfrac{{{V}_{eff}}}{{{R}_{i}}+{{R}_{series}}}\Rightarrow {{I}_{eff}}=\dfrac{112}{0.5+0}\Rightarrow {{I}_{eff}}=\dfrac{112}{0.5}\Rightarrow {{I}_{eff}}=224A. $ This is a dangerously high amount of current, which can seriously damage electrical appliances. Hence, the series resistor acts as a limiting agent to limit the amount of current flowing in the circuit.

Note: Subsequently, if we find out the voltage drop across the series resistance, $ {{V}_{s}}={{I}_{_{eff}}}\times {{R}_{series}}\Rightarrow {{V}_{s}}=7\times 15.5\Rightarrow {{V}_{s}}=108.5V. $
This voltage summing up with the emf of the battery and voltage drop across the internal resistance of the battery is equal to the potential of the external dc source. This proves the Kirchoff’s voltage law.