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Two batteries of emf 4 V and 8 V with in...

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

A

`(1)/(12)` A and `12 V`

B

`(1)/(g) A` and 9 V

C

`(1)/(6)` A and 4 V

D

`(1)/(3) A` and 3 V

Text Solution

Verified by Experts

` I = (8 - 4)/(1 + 2 + 9) = (1)/(3) A , V_(PQ) = (1)/(3) (9) = 3 V `
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