00:01
In this problem, we have been given the following circuit and the batteries are having potential difference of 12 volt as indicated here.
00:09
Also, the batteries do not have any internal resistance as such.
00:13
And the value of the resistors, resistance is given here.
00:17
And we need to first determine the current that's flowing through each resistor.
00:22
So to get the current, let's first say that i -1 current is coming out of this 12 -volt battery on the right.
00:29
I2 current is coming out of this 12 -volt battery on the left -hand side.
00:35
And here we observe at this junction a, the total current which is coming here is i -1 plus i -2.
00:41
And according to kirchhoff's current law, the total current leaving this junction and flowing through r3, that will be i -1 -plus i -2.
00:49
And now let's apply kirchov's voltage law to the two loops here.
00:53
In the first loop, on the right, when we apply kirchov's voltage law, we're going to get to.
00:59
The equation 0 plus 12 minus i 1 r1 minus r3 into i1 plus i2 and that's equal to 0 and let's mark it as equation 1 so further let's simplify this even more by putting the value of the resistances and then we will get another equation that's 12 is equal to i1 into 2 .25 plus r3 which is having the value 7 om times i1 plus i2 so let's separate that so it will be 7 i1 plus 7 i2 and that implies we're going to get the equation here 9 .25 i1 plus 7 i2 is equal to 12 so this is equation 1 also we apply kirchhoff's voltage lot to this loop 2 here and we're going to get the equation 12 minus i2 r2 minus r3 into i1 plus i2 and that's equal to 0 putting the values here once again we're going to get 12 is equal to i2 into r2 which is 3 .35 i2 plus 7 times i 1 plus i 2 so that will be 7 i1 plus 7 i2 and simplifying this even more we're going to get 10 .35 i2 plus 7 i1 is equal to and this is equation 2 and now let's solve this equation so we're going to get the value for i1 and i2 so to solve this let's multiply this equation 1 width so here we will make the coefficient of i2 same so let's multiply this with 10 .35 and multiplying equation 2 with 7 so first we multiply equation 1 with 10 .35 throughout so multiplying 9 .25 with 10 .35, we're going to get 95 .74 i1 plus 7 into 10 .35 i2, and that's equal to 12 times 10 .35.
03:23
So this is updated equation 1.
03:25
And similarly, when we multiply equation 2 with 7, we're going to get 49i1 plus 7 into 10 .30.
03:34
And that's equal to 84 and let's mark it as equation two.
03:41
And now we can subtract equation two from equation one.
03:45
So when we subtract, we observe that the term containing i2, that gets cancelled.
03:50
And we're going to get here 46 .74 i1 and that's equal to 12 times 10 .35 minus 84.
04:02
And that gets us 40 .2.
04:05
So from here when we divide 40 .2 with 46 .74, we're going to get the value for i1 coming out to be 0 .86 ampers...