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Q.
In the circuit shown, the current in the $2\, \Omega$ resistor is _______ $A$.
Current Electricity
Solution:
Applying KVL to loop $APQBA$,
$ 2 I _{1}+\left( I _{1}- I _{2}\right)+3 I _{1}=10 $
$6 I _{1}- I _{2}=10$
$\therefore I _{2}=6 I _{1}-10$
Considering loop $CDQPC$,
$3 I _{2}-\left( I _{1}- I _{2}\right)=6 $
$4 I _{2}- I _{1}=6$
Using equation (i),
$ 4\left(6 I _{1}-10\right)- I _{1}=6$
$ 23 I _{1}-40=6$
$\therefore I _{1}=\frac{46}{23} A $
$ \therefore I _{1}=2 \,A$
