00:01
Here in this given problem there are 5 capacitors arranged in a mixed grouping.
00:09
First of all this is 7 .5 nano farad alone in a branch, the uppermost branch.
00:19
Then in the middle branch 3 capacitors in series and these are 18 .0 nano farad, 30 .0 nano farad and 10 .0 nano farad.
00:35
Then in the lowermost branch again a single capacitor only and this is 6 .5 nano farad.
00:47
The terminals a and b between them the potential applied that is 25 volt.
00:58
In the first part of the problem we have to find equivalent capacitance of this circuit between a and b.
01:05
So first of all as these 18 .0 nano farad, 30 .0 nano farad and 10 .0 nano farad are in series.
01:25
So the net capacitance cs will be given by 1 by cs is equal to 1 by 18 .0 plus 1 by 30 .0 plus 1 by 10 .0.
01:38
Here this lcm that is 90, so this is 5 plus 3 plus 9.
01:49
Means this is 17 by 90 or we can say cs is equal to 90 by 17 nano farad or this is 5 .29 nano farad.
02:08
Now all the three branches are in parallel.
02:12
So net capacitance between them cab this is simply 7 .5 plus 5 .29 plus 6 .5 nano farad means cab that is calculated to be equal to 19 .3 nano farad which is the answer for the first part of this given problem here...