00:01
In this question we are given a circuit in the figure and we want to write the three loop equations based on kerchops rule for the different loops.
00:13
So one of the main things to note is that when we are following the loop, when we bypass a battery, if you are moving from the minus to the plus end, we will add our potential.
00:31
But if you are moving from the plus to the minus m, we'll subtract the potential.
00:38
Next, if you are bypassing a resistor, if a current is in this direction and we are following the same direction as the current, then we will subtract away the potential drop across this resistor, that is ir.
01:00
But if we are moving against the current then we will add the potential drop which is plus i so this is all the all the considerations that we'll need so let us start with loop 1 so for loop 1 we start off with the battery we are moving from minus to positive so we will add our potential that's v1 then we bypass r2 in the same direction as the current so we'll subtract away the potential drop i2 r2 next we are bypassing r1 right in the same direction as the current and we know that because at the junction where there is i2 and i4 coming in we know that from kerchhoff's junction this the left wire must have a current that is flowing out of the junction because you can't have all of the current flowing inwards into a junction right you will not get a net current of zero for that particular junction which you need so the current must be flowing to the left and hence the loop is in the same direction as our current and so we would subtract away the potential drop.
02:45
So we're assuming that the current over here will be equals to i1, make the same current that is flowing outwards from the battery.
03:00
There'll be i1r1.
03:03
We create this to zero because by kerchov's rule we know the entire loop must have a total of zero potential right after adding all of them.
03:16
So this is our first equation.
03:18
So now we move on to the second loop using the same idea.
03:23
We start off at the battery again, v2.
03:27
Now for this time we are moving from positive to negative n of the battery.
03:32
So we have to subtract v2...