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Two batteries of e.m.f. 4V and 8V with i...

Two batteries of e.m.f. `4V` and `8V` with internal resistances `1 Omega` and `2 Omega` are connected in a circuit with a resistance of `9 Omega` as shown in figure. The current and potential difference between the points `P` and `Q`

A

`1/3 A` and 3 V

B

`1/6 A` and 4 V

C

`1/9 A` and 9V

D

`1/2` A and 12 V

Text Solution

Verified by Experts

The correct Answer is:
A
`I = (8-4)/(2 + 1 + 9) = 4/12 = 1/3 A`
`V_(P) = 1 xx 1/3 + 4-8 + 2 xx 1/3 = V_(Q)`
`rArr V_(P) - V_(Q) = 4 -1 = 3V`
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