00:01
Hello students in the question.
00:02
We're given a circuit diagram here.
00:05
We have to find current through the circuit and potential difference across various resistors.
00:12
So here we'll make use of air shaft voltage law, according to which i can write summation of potential difference.
00:29
It is equal to or summation of current into resistance.
00:33
It is equal to 0.
00:35
So here for loop abef, let us assume clockwise direction.
00:45
So here we have according to kirchhoff voltage law equation would be in this case for loop abef.
01:01
We can write equation here.
01:04
We are moving from positive terminal to negative terminal to minus 12 volt again.
01:11
Here we are moving from negative terminal to positive terminal of battery.
01:15
So here positive 18 volt here.
01:18
We are moving in the direction of current through the resistor to minus i1 into 2 ohms.
01:27
It is equal to 0 here.
01:28
We get 6 volt minus 2 times i1.
01:33
It is equal to 0 or we have i1.
01:36
It is equal to 6 divided by 2.
01:39
That is comes out to be equal to 3 amperes.
01:43
Now for loop bct, let us consider the direction of loop to be clockwise again.
01:54
So here we have equation here.
01:56
We are moving from positive terminal of battery to negative terminal to minus 6 volt.
02:02
And here we are moving from negative terminal to positive terminal to plus 12 volt.
02:08
Here we are moving opposite direction to direction of current.
02:12
So positive i3 into 6 ohms.
02:17
It is equal to 0.
02:18
From here we get 6 volt plus 6 i3.
02:23
It is equal to 0 or we get i3.
02:26
It is equal to minus 1 ampere.
02:30
Here negative sign denotes current is opposite in direction to the direction used in the circuit.
03:02
We also have kirchhoff's current law according to which we can write summation of current at the junction.
03:23
It is equal to 0...