Home
Class 12
PHYSICS
Two cells of emf E(1) and E(2) are to ...

Two cells of emf `E_(1) ` and `E_(2)` are to be compared in a potentiometer `(E_(1) gt E_(2))` . When the cells are used in series correctly , the balancing length obtained is 400 cm. When they are used in series but `E_(2)` is connected with reverse polarities, the balancing obtaiend is 200cm. Ratio of emf of cells is

A

`3:2`

B

`3:1`

C

`4:1`

D

`4:3`

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.

Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.

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

In a potentiometer experiment two cells of e.m.f. E1 and E2 are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29 cm . The ratio (E_(1))/(E_(2)) of the e.m.f. of the two cells 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

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) .

In a potentiometer experiment, two cells connected in series get balanced at 9 m length of the wire. Now, if the connections of terminals of cell of lower emf are reversed, then the balancing length is obtained at 3 m. The ratio of emf's of two cells will be

In the above question, if the balancing length for a cell of emf E is 60 cm , the value of E will be

Two cells of emf E_(1) and E_(2) have internal resistance r_(1) and r_(2) . Deduce an expression for equivalent emf of their parallel combination.

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