Question

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

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

Answer 1

Answer: (A)

Applying Kirchhoff's voltage law,

In loop I,
$-27-6I_2-2I_1+24=0$
$6I_2+2I_1=-3\dots \left(i\right)$
In loop II,
$-27-6I_2+4I_1+24=0$
$6I_2-4I_3=-27\dots \left(ii\right)$
At junction $P$, $I_1-I_2-I_3=0\dots \left(iii\right)$
Solving equations $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$ we get
$I_1=3A$, $I_2=-3/2A$, $I_3=9/2A$.