00:01
Hi, in the given problem here this is a circuit diagram having a combination of various resistors.
00:12
The first one is 2 .0 om, then here this is 1 .0, another 1 .0 om in series with it, then 2 4 ome resistors in parallel with each other.
00:45
And then finally, a 3 .0 -oam resistor.
00:55
Then across all of them, a 6 .0 -home resistor, and then a battery, providing a potential of 12 .0 volt.
01:11
Now, in the first part of the problem, we have to find equivalent resistance between the terminals a and b.
01:23
First of all, these 2 4 -oom resistors are in a parallel combination.
01:32
So their net resistance will be denoted by r subscript p.
01:38
Then these 2 1 -oom resistors are in series.
01:45
So we will represent their net combination as r s.
01:50
So using the rule of parallel combination here this rp will be given by 4 .0 by 2 ome means this is 2 .0 and this series combination r s is 1 .0 plus 1 .0 ome means this is 2 .0 om.
02:14
Now this r p and r s are in parallel combination so, the net resistance may be represented as rp dash that will be given by 2 .0 divided by 2.
02:41
As we know, when identical resistors are put in parallel combination, then net resistance of parallel combination is given by the value of one resistor divided by the number of resistors.
02:55
Here the two resistors, rp and rs both are having 2 .0 om and these are two in number.
03:01
That's why this is 1 .0 om.
03:05
Finally, this rp dash complete here, this is rp dash complete.
03:17
This is in series with 2 .0 om and 3 .0 om.
03:21
So, as r p dash 2 .0 om and 3 .0 om are in series.
03:35
So the net resistance which may be represented as r s dash is given as 2 .0 plus 1 .0 for rp dash plus 3 .0 ome which comes out to be 6 .0...