Question

The current I in the 5$\Omega$ resistor for the circuit shown in figure below is_____A.

• A. 3.68

Answer 1

The correct option is A 3.68
At node-1
$10=\frac{V_1}{3}+\frac{V_1-V_2}{2}$
(or) $V_1(\frac{1}{3}+\frac{1}{2})-\frac{V_2}{2}-10=0$
$0.83V_1-0.5V_2-10=0$
At node 2 and 3, the supernode equation is
$\frac{V_2-V_1}{2}+\frac{V_2}{1}+\frac{V_3-10}{5}+\frac{V_3}{2}=0$

or $-\frac{V_1}{2}+V_2[1+\frac{1}{2}]+V_3[\frac{1}{5}+\frac{1}{2}]=2$
$-0.5V_1+1.5V_2+0.7V_3-2=0$
$V_2-V_3=20$
Current in the $5\Omega$ resistor is,
$I_5=\frac{V_3-10}{5}$
$V_3=-8.42V$
$\therefore Current,I_5=\frac{-8.42-10}{5}=-3.68A$
i.e current flows towards node-3