Question

A student is asked to connect $12$ cells each of emf $2\text{V}$ and internal resistance $0.25\Omega$ in series with an external resistance of $3\Omega$ but few cells are connected wrongly by him with its terminal reversed. If the current in the circuit is $2\text{A}$ then the number of cells connected wrongly is-

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Given, $n=12;E=2\text{V};r=0.25\Omega$

$R=3\Omega ;I=2\text{A}$

Net emf of the combination is,

$E^{\prime }=(n-2m)E$

Where ,

$n$ -is total cells ; $m$ -is wrongly connected

Total resistance $R^{\prime }=R+nr$

Current through the circuit is,

$I=\frac{E^{\prime }}{R^{\prime }}=\frac{(n-2m)E}{R+nr}$

$2=\frac{(12-2m)2}{3+(12\times 0.25)}$

$6=12-2m$

$2m=6$

$m=3$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.