00:01
In this question we're given the following circuit and we want to solve for i1, i2 and i3.
00:11
Now one of the first thing that we can do over here is to simplify this circuit a little bit.
00:18
We see that these two resistors are in parallel with each other.
00:26
We can add up to give us an effective resistance.
00:29
So that will be taking 1 over 2r plus 1 over 2r inverse it will be equivalent to just r so what we can do is we will simplify this maybe take away one of the circuit just replace this with r and we will hence know that the current passing through this resistor if effective resistance are is i3.
01:11
And we can form our loop.
01:15
So the loop equations that we're going to have, first loop will be on the left over here.
01:23
In the clockwise manner, the equation that we will have is that, okay, first we start off from passing through the battery for minus to plus, so we add the potential.
01:38
Then we bypass the resistor r on the left in the same direction with the current i2 so we subtract away the potential drop then we bypass the second resistor with current i1 the same direction as the current so we subtract as well equate this to zero the next look that we're going to look at will be on the right hand side we can go in the anticlockwise manner because they will follow the current.
02:13
Alright.
02:17
For the second loop, we are passing the battery from minus to plus again.
02:21
So we add potential.
02:25
Then we bypass the effective resistance r with current i3 in the same direction...