1.

Find the potential differnce between the points A and B and that between E and F of the circuit in. .

Answer»

`V_(AB)=5 V`
`V_(EF)=5 V`
`V_(AB)=0`
`V_(EF)=0`

Solution :The distribution of CHARGE is shown in the figure, in compliance with the POIN RULE. Applying loop rules for loops `A, B`, and `D` with the point rule. Applying loop rules for loop `A,B` and D` we get
.
`(-q_(2))/5+(q_(3))/(0.75)+(q_(1))/(15)=0`
or `q_(1)-3q_(2)+20q_(3)=0`
`-(q_(2)_q_(3))/(15)-q_(3)/(0.75)+(q_(1)-q_(3))/5-q_(3)/(0.75)=0`
or `3q_(1)-q_(2)-44q_(3)=0`
or `23-q_(2)/5-(q_(2)-q_(3))/(15)-q_(2)/5=0`
or `345=7q_(2)+q_(3)`
Solving for `q_(1),q_(3)`, we get
`q_(1)=(19xx345)/(92),q_(2)=(13xx345)/(92),q_(3)=(345)/(92)`
Potential difference berween `A` and `B` is `q_(3)/(0.75)=(345)/(92)xx4/3=5 V`
Potential difference between `E and F` is also `5 V` but in the oposite direction.


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