Question

A storage battery is connected to a charger for charging with a voltage of $12.5\text{V}.$ The internal resistance of the storage battery is $1\Omega .$ When the charging current is $0.5A,$ the emf of the storage battery is

• A. $13\text{V}$
• B. $12.5\text{V}$
• C. $12\text{V}$
• D. $11.5\text{V}$

Answer 1

The correct option is C $12\text{V}$

Let the current $i$ enters the cell through its $+ve-$terminal during the process of charging.


Here $\epsilon$ be the emf of the storage battery,

We can write the equation relating potential of $A$ and $B$ as,

$V_A-\epsilon -ir=V_B$

The storage battery have been connected to an external power supply of $12.5\text{V}$ that means,

$V_A-V_B=12.5\text{V}$

(The direction of current flow is from high potential to low potential)

$\Rightarrow V_A-V_B=\epsilon +ir$

$\text{Or,}\epsilon +ir=12.5$

Given, $i=0.5\text{A},r=1\Omega$

$\Rightarrow \epsilon +(0.5\times 1)=12.5$

$\therefore \epsilon =12.5-0.5=12\text{V}$

Why this question ? $\underset{–}{\text{Tip}}:$Internal resistance of battery means that it has small resistance associated with it, and in such problems we always try to apply the ohm's law while travelling between the two points.