Question

In the circuit shown, the value of currents $I_1$, $I_2$ and $I_3$ are

• A. $3\,A$, $\frac{-3}{2}\,A$, $\frac{9}{2}\,A$
• B. $\frac{9}{2}\,A$, $3\,A$, $-\frac{3}{2}\,A$
• C. $5\,A$, $4\,A$, $-3\,A$
• D. $7\,A$, $\frac{5}{4}\,A$, $\frac{9}{2}\,A$

Answer 1

Answer: (A)

Applying Kirchhoff's voltage law,

In loop I,
$-27-6I_{2}-2I_{1}+24=0$
$6I_{2}+2I_{1}=-3\quad\ldots\left(i\right)$
In loop II,
$-27-6I_{2}+4I_{1}+24=0$
$6I_{2}-4I_{3}=-27\quad\ldots\left(ii\right)$
At junction $P$, $I_{1}-I_{2}-I_{3}=0\quad\ldots\left(iii\right)$
Solving equations $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$ we get
$I_{1}=3\,A$, $I_{2}=-3/2\,A$, $I_{3}=9/2\,A$.