00:01
Hi students, here in the question first of all we need to find the current value.
00:05
Firstly applying kcl at the top node, see the equation becomes i1 equals to i2 plus i3.
00:15
Then applying kvl on the left loop, get pi minus 1 i1 r1 minus pz equals to 0.
00:29
On the right loop we get z minus i3 r3 minus i3 r4 equals to 0.
00:39
Now solving for the currents, for first kvl equation we have i1 equals to pi minus pz upon r1 equals to 13 minus 4 upon 7 which is equals to 1 .286 ampere.
01:04
We have i3 equals to 4 upon 3 that is 4 upon 6 which is equals to 0 .667 ampere.
01:15
Using kcl we have i2 equals to 1 .286 minus 0 .667 which is equals to 0 .619 ampere.
01:30
Solving for i3, i4 and i5 we have the value i4 equals to 0 .667 as i4 equals to i3 and we have i5 equals to 0 .619 ampere because i5 equals to i2.
01:54
Now to find the power supplied, the power supplied is given by the formula power p equals to v into i, putting up the values we have 13 into 1 .286 give us power equals to 16 .718 this is power by next by vz we have p equals to v into i that is 12 into 0 .619 which gives us power equals to 2 .476 volt...