Two cells $A$ and. $B$ are connected in the secondary circuit of a potentiometer one at a time and the…

Two cells $A$ and. $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively $400 \mathrm{~cm}$ and $440 \mathrm{~cm}$. The emf of the cell $A$ is 1.08 volt. The emf of the second cell $B$ in volts is
  1. $1.08$
  2. $1.188$
  3. $11.88$
  4. $12.8$

Solution

$\begin{array}{lc}\frac{E_1}{E_2}=\frac{l_1}{l_2} & \\ \Rightarrow & \frac{1.08}{E_2}=\frac{400}{440} \\ \Rightarrow & E_2=\frac{440 \times 1.08}{400}=1.188 \mathrm{~V}\end{array}$

Asked in: AP EAMCET 2004

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