A wire of resistance `10 Omega` is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a Battery of 3 V and internal resistance `1 Omega` as shown in the figure. The currents in the two parts of the circle are

Options

A. `6/23 A` and `18/23` A
B. `5/26 A` and `15/26` A
C. `4/25` A and `12/25` A
D. `3/25 A` and `9/25 A`

Answer

A

Explanation

`I = 3/(I + (2.5 xx 7.5)/10) = 24/23 A`
`I_(1) = (7.5)/10 xx 24/23 = 18/23 A`
` I - I_(1)`
`I_(2) = 6/23 A`