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

Options

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

Answer

A

Explanation

`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`