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Find the potential differece between the poinst A and B and that between E and F of the circuit shown in. |
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Answer» Left loop: `-q_(1)/5+q_(3)/(0.75)+q_(1)/(15)=0` or `q_(1)-3q_2+20 q_3=0` (i) Middle loop : `-(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` (ii) Outermost loop: `23-q_(2)/5-(q_(2)+q_(3))/(15)-q_(2)/5=0` Outermost loop : `23-q_(2)/5-(q_(2)+q_(3))/(15)-q_(2)/5=0` or `345=7q_(2)+q_(3)` (III) `Solving for `q_(1),q_(2)`, and `q_(3)`. we get `q_(1)=(19xx345)/(92),q_(2)=(13xx345)/(92),q_(3)=(345)/(92)` Therefore, POTENTIAL difference between `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 opposite direction.  .
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