The potential difference between the points `A` and `B` and that between `E` and `F` of the circuit shown in figure respectively are :
.

Options

A. 5V,5V
B. 10V,5V
C. 15 V,5V
D. 0V,0V

Answer

A

Explanation

The distribution of charge is shown in figure according to the junction rule.

Applying loop rules
For loop `-1:-(q_(2))/(5)+(q_(3))/(0.75)+(q_(1))/(15)=0`
`rArr q_(1)-3q_(2)+20 q_(3)=0` ....(1)
For loop `-2 `:
`-(q_(2)+q_(3))/(15)-(q_(3))/(0.75)+(q_(1)-q_(3))/(5)-(q_(3))/(0.75)=0`
`rArr 3q_(1)-q_(2)-44 q_(3) =0` ....(2)
For loop `- 3 : 23-(q_(2))/(5)-(q_(1)+q_(3))/(15)-(q_(2))/(5)=0`
`rArr 345 =7q_(2)+q_(3)` .....(3)
Solving for `q_(1),q_(2),q_(3)`
`q_(1)=(19xx345)/(92),q_(2)=(13 xx345)/(92),q_(3)=(345)/(92)`
`:. p.d` between `A` and `B` = (q_(3))/(0.75)=(345)/(92)xx(4)/(3)=5V`
`p.d` between `E` and `F` is also `5V` but in the opposite direction.