Question

The emf of a storage battery is $90\text{V}$ before charging and $100\text{V}$ after charging. When the charging began the current was $10\text{A}$. What is the current at the end of charging if the internal resistance of the storage battery during the process of charging may be taken as constant and equal to $2\Omega$?
Assume the voltage of the charging point to be constant.

• A. $2\text{A}$
• B. $1.5\text{A}$
• C. $6\text{A}$
• D. $5\text{A}$

Answer 1

The correct option is D $5\text{A}$

When charging began, $i=10\text{A}$ in circuit.

Emf of battery, $E_1=90\text{V}$


Let the potential difference at which the charging point is operating be $V\text{Volts}$.

Using KVL in the loop, we get

$-E_1-ir+V=0$

$\Rightarrow -90-10\left(2\right)+V=0$

$\Rightarrow V=110\text{Volt}$

When charging just completed let current be $i_2$ in circuit.


Again apply the KVL in the loop during charging,

$V=E_2+i_{2}r$

$\Rightarrow 110=100+i_2\left(2\right)$

$\Rightarrow 2i_2=10$

$\therefore i_2=5\text{A}$

Hence, option $\left(d\right)$ is correct.
$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: If the same charging point is used}\\ \text{throughout for charging a storage battery,}\\ \text{the potential difference across ends of}\\ \text{battery will be equal to operating voltage}\\ \text{of charging plant, provided that if it is}\\ \text{operating at steady state.}\end{array}$$