00:01
In this problem, we have been given a part of the circuit and we need to determine the equivalent resistance between the points a and b.
00:09
So here the resistances of the resistors are given and we can observe that.
00:14
First, these two resistors, they are in series.
00:18
So their equivalent value will be 6 om.
00:22
And now this 6 om and this 4 -oom resistors, they are in parallel.
00:26
So their equivalent resistance will be six times six divided by, so it will be four because this is 4 oom and when we add them together it will become 6 o.
00:37
So 6 times 4 divided by 6 plus 4 and that will be 2 .4 oms.
00:43
And now this total resistor combinations can be replaced by 2 .4 oom resistor and that's in series with this 6 oom.
00:52
So the circuit simplifies to just 8 .4 oom resistor between the upper branch and contained in between the points a and b.
01:02
Now similarly, we simplify lower branch.
01:06
So these two resistors, they are in parallel, and their equivalent resistance will be 8 times 8 divided by 8 plus 8.
01:13
So that comes out to be 4 .m.
01:15
And this 4 oom is in series with this 4 oom resistor, and the total resistance now it will be, 8 o.
01:23
So we have just simply used the formula to compute resistance in parallel as r1 times r2 divided by r1 plus r2 and resistance in series given by r1 plus r2.
01:37
And now we can see that again these two resistors, they are in parallel.
01:41
So the equivalent resistance between a, b, that can be given as 8 times 8 .4 divided by 8 plus 8 .4.
01:50
So when we multiply 8, with 8 .4 and divided by 8 plus 8 .4.
01:57
We get the total resistance as 4 .1 oom approximately.
02:07
And now considering the potential difference applied across the points a and b as 12 volt.
02:14
We need to determine the current across each resistor...
