two cells of e.m.f E1 and E2 (E1>E2) are connected as shown in the fig

1.	two cells of e.m.f E1 and E2 (E1>E2) 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 , calculate the ratio of E1 and E2[3:2]
Answer» E1=k×300 For the second part it depends on the connection of cell if NEGATIVE terminal on same side then : E1+E2=k×100 If it travels in the OPPOSITE direction then : E1-E2=k×100 So, the question here is INCOMPLETE and this is all the DATA that can be concluded.

two cells of e.m.f E1 and E2 (E1>E2) 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 , calculate the ratio of E1 and E2[3:2]

Answer»

E1=k×300

For the second part it depends on the connection of cell if NEGATIVE terminal on same side then :

E1+E2=k×100

If it travels in the OPPOSITE direction then :

E1-E2=k×100

So, the question here is INCOMPLETE and this is all the DATA that can be concluded.

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