Question

Two cells of emfs, E1 & E2 (E1 < E2) are connected as shown in figure



When a potentiometer is connected between A & B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A & C, the balancing length is 100cm. Calculate the ratio of E1 & E2.

Answer 1

the answer is 3:1.Because,the formula is E=Irl

Answer 2

Answer:

Balancing length of the cell 1 = 300 cm

Balancing length of the cell 2 = 100 cm

WKt,

The emf's of the cells are proportional to its balancing lengths

Hence,

            $\frac { { E }_{ 1 } }{ { E }_{ 2 } } \quad =\quad \frac { L1 }{ L2 }\\\frac { { E }_{ 1 } }{ { E }_{ 2 } } \quad =\quad \frac { 300 }{ 100 }\\ \frac { { E }_{ 1 } }{ { E }_{ 2 } } \quad =\quad \frac { 3 }{ 1 } = 3:1$ which is the required ratio.