Question

In the circuit shown, the current in the $2\Omega$ resistor is _______ $A$.

Answer 1

Answer: 2

Applying KVL to loop $APQBA$,
$2I_1+\left(I_1-I_2\right)+3I_1=10$
$6I_1-I_2=10$
$\therefore I_2=6I_1-10$
Considering loop $CDQPC$,
$3I_2-\left(I_1-I_2\right)=6$
$4I_2-I_1=6$
Using equation (i),
$4\left(6I_1-10\right)-I_1=6$
$23I_1-40=6$
$\therefore I_1=\frac{46}{23}A$
$\therefore I_1=2A$