00:01
Hi, in this question we have this circuit in which r1 is given as twice of r, r2 is given as r and r3 is also given as r.
00:13
So we have to calculate the magnitude of current passing through resistor r2 in terms of v by r.
00:21
So now if we see in this circuit and if we label it as a, b, c, d, t and f, and if we look at this as loop 1 and if we take this as loop 2 and in this one we have current i1 flowing and in this one we have current i2 flowing.
00:53
So now if we apply in loop a, b, c, d, a, we can write minus 2v minus r1 i1 plus v minus r3 into i1 minus i2 plus v equals to 0.
01:18
So now from here we can write minus twice of r i1 minus r1 minus i1 minus i2 equals to 0.
01:34
So we have like put the value of r1 and r3 from the above three equations.
01:40
So now on solving for i1 we get i1 equals to 1 upon 3 times i2.
01:51
Now we'll apply a loop in the next part that is dcfed.
01:57
So from there we have minus v minus r3 into i2 minus i1 minus v minus r2 i2 is equals to 0...