00:01
In the given problem, there are various resistors joined in mixed grouping.
00:10
First of all here this is r1 which is in series with r2, then two resistors are in parallel with each other and this parallel combination is in series with r1 and r2.
00:25
The upper one are 3, the lower one are 4.
00:29
And then one more parallel combination of two resistors and this parallel combination is in series with the other one upper one r5 lower one r6 and finally a source of voltage and circuit is closed this voltage applied is having a value 20 .0 1 now the values of resistances r1 is 5 .00 om.
01:09
R2 is 10 .0 om.
01:13
R3 is again 5 .00 om.
01:19
R4, 5 .00 om, r5 .00 om, r5, 2 .00 om.
01:30
And finally r5.
01:31
This is also 2 .00 oom now in the first part of the problem after finding the equivalent resistance of this combination we have to find the potential drops across each and every resistance so first of all there are two parallel combination first of all this r3 with r4 and r5 with r6 the net resistances are assumed to be if we consider the terminal terminals here this is a b c and the mid one d and finally this is e.
02:20
So the resistance of this parallel combination will be named as r -c -d and here this last one will be named as r -d -d.
02:37
Now as r3 and r4 are in parallel and having identical resistor each having a value of 5 om therefore the net combination parallel combination rcd will be given by 5 .00 by 2 om means this is 2 .50 om similarly as rr as r as r as r as r r5 and r6 are in parallel and they are also having identical resistor each having same value of 2 .00 om so their parallel combination will come out to be rde is equal to 2 .00 om divided by 2 which comes out to be 1 .00 om.
03:40
Finally is r1, r2, rcd and rde all 4.
03:46
Are in series...