Question

Two cells of different emf and internal resistances connected in series with an external resistor. The current in the circuit is $3.0\text{A}$. When the polarity of one cell is reversed, the current in the circuit becomes $1.0\text{A}$. The ratio of the emf of the two cells is

• A. $3:2$
• B. $3:1$
• C. $2:5$
• D. $2:1$

Answer 1

The correct option is D $2:1$

Let
$E_1,E_2$ be emf of cells,
$r_1,r_2$ be internal resistance of cells
$I_1$ be current in circuit when cells are connected in same polarity
$I_2$ be current in circuit when cells are sonnected in opposite polarity
When connected in same polarity
$E_1+E_2=I_1[R+r_1+r_2]....\left(1\right)$
When connected in opposite polarity
$E_1-E_2=I_2[R+r_1+r_2]....\left(2\right)$

Using equations $\left(1\right)\&\left(2\right)$
$\Rightarrow \frac{E_1+E_2}{E_1-E_2}=\frac{I_1}{I_2}\Rightarrow \frac{E_1+E_2}{E_1-E_2}=\frac{3}{1}$
$\Rightarrow \frac{E_1}{E_2}=\frac{4}{2}=\frac{2}{1}$
$\begin{array}{c}{E}_1:E_2=2:1\end{array}$