Question

Two cells of e.m.f. $E_1$ and $E_2$ are joined in series and balancing length of the potentiometer wire is 625 cm. If the terminals of $E_1$ are reversed, the balancing length obtained is 125 cm. Given $E_2>E_1$, the ratio $E_1:E_2$ is ?

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

Answer 1

The correct option is C $3:2$

We know that in potentiometer voltage is directly proportional to balancing length

if both cell with same polarity connected in series then,

$E_1+E_2=625$

if both cell with opposite polarity connected in series then,

$E_1-E_2=125$

By dividing both the equation

$\frac{E_1+E_2}{E_1-E_2}=\frac{625}{125}=5E_1+E_2=5(E_1-E_2)=5E_1-5E_{2}4E_1=6E_2\frac{E_1}{E_2}=\frac{6}{4}=\frac{3}{2}$

Hence,

option(C) is correct option