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Two cells are connected in series and e.m.f. of combination is found to balance against length of 450 cm of potentiometer wire. Whent the two cells are connected in parallel emf of combination balances against 50 cm length of wire. The ratio of e.m.f. of two cells is

A

2.25

B

1.55

C

1.25

D

2.55

Text Solution

Verified by Experts

The correct Answer is:
C
`(E_(1))/(E_(2))=(l_(1)+l_(2))/(l_(1)-l_(2))=(450+50)/(450-50)=5/4=1.25`
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