00:01
In the first part of this problem, we are going to calculate the current passing through each of the u1 resistors, that is i1, i2, and let's name the loop of the u1 circuits from left to right as 1 and 2.
00:15
Then using kerkhov's current rule, we can write i3 equals to i2 minus i1.
00:21
Let's call it equation a.
00:23
Now let's apply the karchov's voltage rule in loop 1, e1 minus 1.
00:30
E2 minus i2 r2 minus i1 i1 equals to 0 let's try the values for this equation 70 .0 volt minus 60 .0 volt minus i 2 into 3 .00 kilo minus i 1 into 2 .00 kilo arm equals to 0 so this can be simplified as 10 .0 volt minus into 3 .00 kilo arm into i2 minus 2 .0 kilo arm into i 2 .0 kilo om into i1 equals to 0 let's call it equation 1 similarly applying the same rule in loop 2 we can write e 3 minus i 3 r3 minus e 2 minus i 2 r2 equals to 0 now let's write the values for this equation 80 .0 volt minus i 3 into 4 .00 kilo om minus 60 .0 volt minus i 2 into 3 .00 kilo om equals to 0 so this equation can be simplified is 20 .0 volt minus into 4 .00 kilo -oam into i3 minus 3 .00 kiloam i2 equals to 0.
01:43
Let's call it equation 2.
01:45
Now setting the value of this i3 from equation a into equation 2, we will get 20 minus 20 .0 volt minus into 7 .00 kiloam i2 plus 4 .0.
02:00
0 .00 kilo -oam i1 equals to 0 let's call it equation 3 now multiplying equation 1 by 2 and then adding the oct -hand equation with equation number 3 we will get 40 .0 volt minus into 13 .0 kiloam into i 2 equals to 0 so this will give us the value for i 2 as i 2 equals to 3 .08 millie ampair.
02:30
Now setting the value of this i2 into equation 1 we will get 10 .0 volt minus 3 .00 kilo -oom into 3 .08 mill ampere into minus 2 .00 kilo -oom into i1 equals to zero.
02:48
Now solving this equation for i1 we will get i1 equals to 0 .385 milli -ampr...