Two primary cells of emfs `E_(1)` and `E_(2)` are connected to the potentiometer wire AB as shown in the figure . If the balancing lengths for the two combinations of the cells are 250 cm and 400 cm , find the ratio of `E_(1)` and `E_(2)` .

Two cells of emfs E_(1) and E_(2) are connected together in two ways shown here. The 'balance points' in a given potentiometer experiment for these two combinations of cells are found to be at 351.0 cm and 70.2 cm respectively. Calculate the ratio of the emfs of the two cells.

Two cells of emf E_(1) and E_(2) ( E_(1) gt E_(2)) are connected in series to potentiometer for balancing length 625 cm . When polarity of E_(2) is reversed then balancing length becomes 125 cm. Then the ratio (E_(1))/(E_(2)) is

Tow cells of e.m.f E_(1) and E_(2) are joined in series and the balancing length of the potentiometer wire is 625 cm. If the terminals of E_(1) are reversed , the balancing length obtained is 125 cm. Given E_(2) gt E_(1) ,the ratio E_(1) : E_(2) willbe

Tow cells of e.m.f E_(1) and E_(2) are joined in series and the balancing length of the potentiometer wire is 625 cm. If the terminals of E_(1) are reversed , the balancing length obtained is 125 cm. Given E_(2) gt E_(1) ,the ratio E_(1) : E_(2) willbe

Two cells of emfs E_(1) and (E_(2) (E_(1) gt E_(2)) are connected as shows 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 E_(1)//E_(2)

Two cells of emf E_(1) and E_(2) are connected to two resistors R_(1) and R_(2) as shown. If E_(2) is short circuited then current through R_(1) and R_(2) are

Two cells of emfs E_(1) and (E_(2) (E_(1) gt E_(2)) are connected as shows in Fig. 6.45 . 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 E_(1)//E_(2) is