00:02
Students, in this question we need to find the value of currents that is i1, i2, i3, i4 and i5, and also the power supplied by the voltage source, power supplied by the voltage source and also power dissipated.
00:22
This is power supplied ps and also power dissipated pd.
00:28
We need to find these things.
00:31
Now, let us apply loop law in the loop 1 here.
00:42
This is one loop, this is one loop and this is another loop.
00:49
Now, applying loop law in loop 1 in this.
00:54
Starting from this, we get loop 1.
01:06
This is loop 1.
01:10
Minus 3 i2 minus 3 i 2 plus 13 minus 5 i 1 equals to 0 by this we get the equation that is 5 i1 plus 3 i2 equals to 13 this is equation 1 now from loop 2 i 3 that is minus 2 i 3 plus 5 i 1 minus 5 i 1 equals to to 0, which can be written as 5 i1 minus 2 i3 equals to 5.
02:01
And this is equation 2.
02:05
In loop 3, in loop 3, starting from the position, we get 5 minus 2 into i4, which is equals to 0, and which gives rise to i4 is equal to 2 .5 amps.
02:33
At the junction, that is here, let us name this as m and name this as m.
02:44
At this junction, we can write i1 plus i3 is equals to i2.
02:59
Therefore, 5 into i1 plus 3 into i1 plus i3 is equals to 13.
03:10
This is from equation 1, which gives rise to 8 i1 plus 3...