00:01
Consider the two loop circuit as shown below.
00:03
Please simplify your answers to simple fractions if necessary and place your answers in the box provided with units.
00:11
Find the current scyone i -2 -i -2 in the circuit when the e -1 value of e -1 and e2 is given for the ideal batteries and r1 and r2 is also given to us.
00:24
So we have to use the loops a -b -e -f -a and e -b -c -d -e in your solution.
00:31
So where is the loop a -b -e -f -a? a -b -e -f -a is this particular loop.
00:37
Let's call it loop 1.
00:39
And e -b -c -d -a.
00:41
So e -b -c -d -a is this loop.
00:44
Let's call it loop two.
00:46
Okay, all right.
00:47
So we are going to use the junction route, kirchov's current law and kirchow's voltage law over here.
00:53
So the junction which we are going to use is b.
00:56
So if you apply k -c -l at the point b, apply kcl at b then the incoming current i1 should be equal to the outgoing current that's call it equation 1 if you apply kvl at in loop 1 then we are going to have uh you're going to start with a right since this loop starts from a so you're going to start with a and move clockwise so that's minus i1 r1 minus i2 r2 then we have a battery minus e2 minus because i1 is going in this direction minus i1 r1 plus e1 is equal to zero so let's substitute all the values so let's write even plus and even as given as 18 more so that's 18 minus e2 e2 is given as two volts minus i1 r1 and i1 r1 becomes two times i1 r1 and this is minus i2 r2 this is equal to zero 18 minus 2 is 16 so so 16 minus 2 times r1.
02:02
What is r1? r1 is 1.
02:04
So 2 times 1 is 2.
02:06
So that's 2 i1.
02:07
And r2 is 3.
02:09
So that's 3 i2.
02:11
That is equal to 0.
02:13
So this becomes 16 is equal to 2i1 plus 3i2.
02:17
And if you flip the equation, then 2i1 plus 3i2 is equal to 16.
02:22
Let's call it equation number 2.
02:24
Then let me talk about kvl in the loop 2.
02:28
So it starts with e.
02:30
So i'm going to also start with e and move clockwise.
02:34
So e2 comes out as this becomes e2, starts with e2, then minus plus, because we are going against the currents, plus i2 r2 minus i3r1, minus e2 minus e3r1 is equal to 0.
02:54
All right, okay.
02:55
So let's try to simplify this and also substitute the value.
02:58
So here we have, okay, that's interesting because both are e2 over here.
03:06
So both actually cancels out each other.
03:09
So these two are canceled.
03:11
And we are left with i2r2.
03:14
Now what is the value of r2? that is three.
03:18
So this is three i2.
03:20
And i3r3 and i3 r3 becomes twice of i3.
03:23
Sorry, i3r1 and i3r1 because twice of i3r1, i3r1 is equal to zero.
03:28
And r1 is nothing but one ome so this becomes three i2 is equal to two i3 right because r1 is just one let's call it equation number three so uh from equation one and two let's try to eliminate the value of i i i'm going to do it over here from equation number one and two i'm going to substitute the value of i1 over here so this becomes two times i2 plus i3 plus three times i2 is equal to six this means that 2 i2 plus 2 i 3 plus 3 i2 is equal to 16.
04:06
2 i2 plus 3 i 2 i 2 is 5 i 2 plus 2 i 3 is equal to i 3 is equal to i 3...