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Two cells of emf E(1) and E(2) are joine...

Two cells of emf `E_(1)` and `E_(2)` are joined in opposition (such that `E_(1) gt E_(2)`) . If `r_(1)` be the internal resistances and R be the external resistance, then the terminal potential difference is

A

`(E_(1) +E_(2))/(r_(1) +r_(2) + R) xx R`

B

`(E_(1) - E_(2))/(r_(1)+r_(2) + R) xx R`

C

`(E_(1) + E_(2))/(r_(1)+ r_(2)) xx R`

D

`(E_(1) -E_(2))/(r_(1) +r_(2)) xx R`

Text Solution

Verified by Experts

The correct Answer is:
B
`I =(E_(1) - E_(2))/(r_(1) + r_(2) + R)`
`V = IR =((E_(1)-E_(2))R)/(r_(1) + r_(2) + R)`
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