Question

Question 3.9: Determine the current in each branch of the network shown in fig 3.30:

Class 12 - Physics - Current Electricity Current Electricity Page-128

Answer 1

Let us first distribute the current in different branches . Here you see there are three loops.
now, equations for different loops using Kirchoff's 2nd law,

$\bf{loop-I}$

$\Sigma E=\Sigma IR\\10I_1+5I_g-5I_2=0\\or,2I_1+I_g-I_2=0-----(1)$

$\bf{loop-II}\\\\\sum E=\sum IR\\5I_g+10[I_2+I_g]-5[I_1-I_g]=0\\10I_2+20I_g-5I_1=0\\or,2I_2+4I_g-I_1=0----(2)$

$\bf{loop-III}\\\\\sum E=\sum IR\\5I_2+10(I_2+I_g)+10I=10\\15I_2+10I_g+10I=10\\or,3I_2+2I_g+2I=2------(3)$

solving equations (1) and (2),
$2I_1+I_g-I_2+2[-I_1+4I_g+2I_2]=0\\9I_g+3I_2=0\\or,I_2=-3I_g----(4)$

in the loop ABCDA,
$10I_1+5[I_1-I_g]-10[I_2+I_g]-5I_2=0\\15I_1-15I_2-15I_g=0\\or,I_1-I_2-I_g=0----(5)$

solving equations (2) and (5),
$2(I_1-I_g)+4I_g-I_1=0\\or,I_1=-2I_g---(6)$

now, using the results of (4) and (6) in equation (3),
$3(-3I_g)+2I_g+2I=2\\2I-7I_g=2----(7)$

using Kirchoff's law,
$I=I_1+I_2=-2I_g-3I_g=-5I_g\\I=-5I_g$
so, equation (7),
$-2(-5I_g)-7I_g=2\\-17I_g=2\\so, finally\:I_g=-2/17A$
so, $I=-5I_g=+10/17A\\\\similarly, I_1=4/17A,and,I_2=6/17A$

so, current in branch AB = 4/17 A
current in branch AD = 6/17 A
current in branch BD = -2/17 A
current in branch BC = 6/17 A
current in branch DC = 4/17 A

Answer 2

The answer is well explained and short with just two equations. See attachment