Question

Two cells of ends $E_{1}and{E}_2(E_1>E_2)$ are connected as shown In 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 l is $\frac{E_1}{E_2}$ is

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

Answer 1

Answer: (D)

3 : 2

Since the length is proportion to Resistance

let us assume $R=cl$ , length

Since it is balanced at 300cm

When connected to AB

$E_1=i{R}_1=ic\times 300$ ...(1)

When connected to AC

$netEMF$ =$E_1-E_2$

$E_1-E_2=ic\times 100$ ...(2)

from equation 1 and 2

$E_2=ic\times 200$

$\frac{E_1}{E_2}=\frac{3}{2}$