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Two sources of emf 6 V and internal resi...

Two sources of emf `6 V` and internal resistance `3 Omega` and `2 Omega` are connected to anexternal resistance `R` as shown. If potential difference across battery `A` is zero, then the value of `R` ( in `Omega`) is

A

`1 Omega`

B

`2 Omega`

C

`3 Omega`

D

`4 Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) `I = (12)/(5 + R) implies V = E - Ir = 0`
`6 - ((12)/(5 + R)) xx 3 = 0 implies R = 1 Omega`
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