00:01
In this problem, we have been given the following circuit in which we have two batteries with potential difference, va and vp, and the resistors, with their values indicated, they are as shown, and we have to determine the current that's flowing through this resistor r2, that's i2, and we have to get this current in terms of the given parameters.
00:23
So first, we take the current that's coming out from this battery with potential difference, ve, let's take it as i1.
00:30
And using kirchav's junction rule at this junction, the current that's coming to this junction is i1, and the current that will be leaving of this junction, that will be equal to the current that's entering or coming to this junction.
00:44
So i1 will be i2 plus x, and that implies the current that's flowing through this or from this battery vb, that is i1 minus i2.
00:56
And here we make use of kirchap's voltage rule, according to that, the summation of potential difference in a loop that is zero so we start at this point let's see so first we select this loop and we move in clockwise manner and we get one equation so let's take the potential at this point a zero volt and the potential at this point that will increase by we amount because we are moving from negative to the positive terminal and the potential at this point that will drop by the amount v equals i r so this we represents the potential drop according to oms law so that will be dropped by minus i1 r1 and similarly the potential at this point that will drop by i2 times r2 and this is equal to zero so from here let's get the value for i1 which we will substitute by using kirch's voltage law in another loop so so we get i1 here as va minus i2r2 over r1.
02:03
Let's mark it as equation 1.
02:06
And now we move in another loop.
02:08
So first we are talking about this loop and we are moving in clockwise manner here as well.
02:15
And according to gitchf's voltage law, let's say we start at this point with zero potential.
02:20
So potential at this point will increase by wavy amount and potential at this point it will decrease by i1 minus i2 times r3 and potential here it will increase because we are moving in a direction opposite to that of the current so that will be plus i2 into r2 and that's equal to zero and from here let's substitute the value of i1 so we're going to get here we be minus let's put the value for i1 from equation 1 so we get we a minus i 2 r2 over r1 minus i 2 times r 3 plus i 2 r2 it's equal to 0 so let's just simplify this to get the value for i 2 in terms of the given parameters so we get v b minus we a r 3 over r1 plus i 2 r2 r3 over r1 plus i 2 r2 r2 r3 over that's equal to zero.
03:35
And from here we can just simplify it even more.
03:40
So there will be one more term to this because we also have this term...