Question

The figure shows a network of resistors and a battery. If $1\text{A}$ current flows through the branch $CF$ then


The emf $E$ of the battery is

• A. $6\text{V}$
• B. $18\text{V}$
• C. $12\text{V}$
• D. $24\text{V}$

Answer 1

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


$V_{\text{CF}}=1\times 2=2\text{V}$

$V_{\text{CE}}=\left(i_1\right)\times 2$

$2=i_1\times 2$

$i_1=1\text{A}$


The P.D across BG $=(2\times 2)+2$

$=4+2$

$=6\text{V}$

Current through BG is,

$i_2=\frac{6}{6}=1\text{A}$

$\therefore$ Total current $I=1+2=3\text{A}$

Total resistance $R^{\prime }=2+\frac{3\times 6}{9}=4\text{A}$

$\therefore E=R^{\prime }I=4\times 3$

$\therefore E=12\text{V}$