Question

Two cells of e.m.f. $E_1$ and $E_2(E_1>E_2)$ are connected as shown in the figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio $\frac{E_1}{E_2}$ is

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

Answer 1

The correct option is D 3 : 2

The balancing length of the potentiometer connected between A and B is given by
$l_{AB}=300cm$
The balancing length of the potentiometer connected between A and C is given by
$l_{AC}=100cm$
The principal of potentiometer states that emf is directly proportional to the balancing length
$E\propto l$

$⟹E_1=K\times 300$.........(i)
(where K is constant of proportionality)
and $E_1-E_2=K\times 100$......(ii)
On solving (i) and (ii) we get
$\frac{E_1}{E_2}=\frac{3}{2}$