By constructing Millman’s equivalent voltage source at the left of terminals a and b in the given circuit, Find current I.

Solution:

Given circuit. Can be drawn as,

Milliman's equivalent voltage source circuit. is,

$R_m \sum_m=I \sum_i^a d 1000 \Omega$....(1)

$\therefore I=\frac{V_m}{R_m+1000}$

Where,

$V_m=\frac{V_1 G_1+V_2 G_2+V_3 G_3}{G_1+G_2+G_3}$ ....... (2)

&

$\text R_m=\frac{1}{G_m}=\frac{1}{G_1+G_2+G_3}\\ $

$\begin{aligned}\\ &G_1=\frac{1}{R_1}=\frac{1}{10}=0.1 ; G_2=\frac{1}{R_2}=\frac{1}{100}=0.01 \mathrm{v} \\ &G_3=\frac{1}{R_3}=\frac{1}{100}=0.01 \mathrm{v}\\ \end{aligned}\\ $

from..... (2)

$\begin{aligned}\\ R_m=\frac{1}{G_m}=& \frac{1}{0.1+0.01+0.01}=8.333 \Omega \\ & R_m=8.333 \Omega\\ \end{aligned}\\ $

from .....(2)

$\begin{aligned}\\ &V_m=\frac{10(0.1)-20(0.01)+50(0.01)}{0.1+0.01+0.01} \\\\ &V_m=10.833 \text { Volts } \\ \end{aligned} \\\\ $

from .....(1)

$\begin{aligned}\\ &I=\frac{10.833}{8.333+1000} \\\\ &I=0.0107 \mathrm{Amp}\\\\ \end{aligned}\\\\ $