00:01
In this circuit we need to find the value of i1, i2, i3, i4, i5 and i6.
00:09
I solve this problem by using mass analysis and in this circuit there are three mass.
00:16
This is mass a.
00:18
So in mass a, i assume that current flowing is ia.
00:23
This is mass two and i assume that in mass two current following is ib and i assume that in mass two current following is ib and and this is mes 3 and i assume that in mes 3 current flowing is ic.
00:37
Now first i calculate the value of ia, i b and ic then by using this current i find the value of i2, i3, i5 and i6.
00:48
Now first i apply kitchoff voltage low in mesa.
00:54
So i start from 14 volt voltage source.
00:58
So current flowing in this direction, ia current flowing in this direction.
01:04
So voltage rise take place.
01:06
So this is plus 14.
01:08
We know register always offer voltage drop.
01:12
So negative 3 into ia minus the current flowing in register r2 is equal to ia minus ib.
01:24
So minus r2.
01:25
R2 is equal to 2 into ia minus i b is equal to 0 after simplification this will become 14 is equal to 3 i a plus 2 i a minus 2 i b and after further simplification this equation will become 5 i a minus 2 i b is equal to 14 now let us assume that this is the equation number first.
02:03
Now next we apply kitchoff voltage low in mass b.
02:07
In mesb current flowing in anti -clockwise direction.
02:12
I start kvl from 25 voltage source...