Question

Two cells of emf $ 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 300cm .
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

Answer: (D)

When potentiometer is connected between A and B, then it measures only $ E_1 $ and when connected between A and C,then it measures $ E_1 - E_2 $ .
$ \therefore \, \, \, \frac{E_1}{E_1 - E_2} = \frac{l_1}{l_2} $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{E_1 - E_2}{E_1} = \frac{l_2}{l_1} $
$ \Rightarrow 1 - \frac{E_2}{E_1} = \frac{100}{300} $
$ \Rightarrow \frac{E_2}{E_1} = 1 - \frac{1}{3} $
$ \Rightarrow \frac{E_2}{E_1} = \frac{2}{3} $
$ \Rightarrow \frac{E_1}{E_2} = \frac{3}{2} $