## In This post you will get answer of Q.

In the circuit shown, current flowing through $25, V$ cell is

Solution:

## Applying $KVL$ in loop $ABCDA, ABFEA, ABGHA$ and $ABJIA$,

we get

$30- i _{1} times 11=-25 ldots$ (i)

$20+i_{2} times 5=25 $… (ii)

$5-i_{3} times 10=-25 ldots$ (iii)

$10+i_{4} times 5=25 $… (iv)

Solving equations (i), (ii), (iii) and (iv) we get

$i _{1}=5 ,A , i _{2}=1 ,A , i _{3}=3, A$ and $i _{4}=3, A$

Hence, current flowing through $25 ,V$ cell is $12, A$.