00:01
Hi, here in this given problem first of all there is a grouping of resistors in which this 10 .0 ome resistance is put in series with the cell providing 25 .0 volt.
00:20
Then here this is another 10 .0 ome which is in parallel with this 5 .00 oom and between the terminals a and b, two more registers which themselves are in series, but they are in parallel with 10 and 5 om.
00:59
And these resistors are 5 .00 om, and this is r whose value is given as 38 .0.
01:10
So actually there are three branches in parallel 10, 0 5 om and 5 plus 38 means 43 om.
01:23
These three are in parallel.
01:26
So in the first part of the problem we have to find current passing through this 38 ome resistance.
01:36
Suppose this is i -dash and the total current in the circuit? suppose that is i.
01:42
So, rs means 5 om and 38 -oam are in series.
01:50
So that comes out to be equal to 5 .00 plus 38 .0 om and this is 43 .0 om.
02:01
So now these three are in parallel.
02:07
So net resistance of this parallel combination 1 by 10 plus 1 by 5, 1 by r p is equal to 1 by 10 plus 1 by 5 plus 1 by now, if you take this 10 multiplied by 43 as an lcm, so here it is 43 plus here it is 43 multiplied by 2 and here it is just 10...