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Question: There is a fixed potential difference between two ends of a potentiometer. Two cells are connected i...

There is a fixed potential difference between two ends of a potentiometer. Two cells are connected in series in such a way that in one arrangement they help each-other whereas in second arrangement they oppose each-other. The balance points for these two arrangements are obtained at 120 cm and 60 cm length respectively. The ratio of emf of two cells is-

A

2 : 1

B

3 : 1

C

1 : 1

D

4 : 1

Answer

3 : 1

Explanation

Solution

ε1+ε2ε1+ε2\frac { \varepsilon _ { 1 } + \varepsilon _ { 2 } } { \varepsilon _ { 1 } + \varepsilon _ { 2 } } = 12060\frac { 120 } { 60 }

ε1 + ε2 = 2ε1 – 2ε­2

ε1 = 3ε2

∴ ε1ε2\frac { \varepsilon _ { 1 } } { \varepsilon _ { 2 } } = 31\frac { 3 } { 1 }