00:01
In this question, we are asked to find the total energy stored in this network and then we are asked to find the energy stored in the 4 .8 micro -farrad capacitor.
00:15
So let's start to solve this problem.
00:17
To solve this problem, we use this formula.
00:19
We know energy stored in a capacitor is given by half c .v.
00:25
Here, v is the voltage across the capacitor.
00:27
Now here in part a we are asked to find the total energy stored in this network.
00:33
Now assume that the total capacitance of this network is equal to c equivalent and the voltage across c equivalent is equal to vab.
00:43
So now energy stored in this network is given by u equals to half c equivalent into vab square.
00:52
Here given that vab is equal to 15 volt.
00:56
So now our next task is to find that.
00:58
The value of c equivalent.
01:00
Here c equivalent is the total capacitance of this network or the capacitance across terminal a and b.
01:05
Now look at the circuit here 6 .2 microferred and 11 .8 microferred capacitors are connected in series.
01:14
So their series combination is given by 6 .2 into 10 to the power negative 6 into 11 .8 into 10 to the power negative 6 over 6 .2 into 10 to the power negative 6 over 6.
01:28
Negative 6 plus 11 .8 into 10 to the power negative 6 and by simplifying this this will become because 4 .06 4 .4 micro ferret now in this network 4 .0 644 microferred capacitor and 3 .50 microferret capacitors are connected in parallel so their parallel combination is given by 4 .0644 into 10 to the power negative 6 plus 3 .5 into 10 to the power negative 6.
02:05
And by simplifying this, this will become equals to 7 .5644 microferred.
02:15
Now, in this network, 8 .6 microferred capacitor, 4 .8 microferate capacitor, and 7 .5 .5 .5 .5 .5 .5 .5 .5.
02:27
5644 microferet capacitor are connected in series.
02:33
So now this circuit looks like this.
02:37
Now here in this network 8 .6 microferred and 7 .5644 microferred capacitor are connected in series.
02:47
So their series combination is given by 8 .6 into 7 .5644 into 10 to the power negative 6 .6.
02:58
Over 8 .6 plus 7 .5644 and by simplifying this this will become equals to 4 .0245 micro ferret so now in the circuit 4 .0245 microferred capacitor and 4 .80 microfair capacitor are connected in series so now this circuit looks like this terminal a 4 .0.
03:33
0 .245 micro ferret capacitor and then we have 4 .8 micro ferret capacitor...