00:01
In this question, we are given this circuit.
00:03
We need to find a total energy stored in the system, energy stored by each capacitor, and compare the sum of answers to the result in part a.
00:13
So to do this question, so i'm going to call this, i'm going to assign c1 is 3 micro -farrant, c -2 is 6 micro -faird, c -3 is 2 micro -faird, c4, c4, c4, 4 is 4 micro ferret.
00:40
And then c1 and c2 are in series.
00:47
So we can calculate c12 to be 1 over c1 plus 1 over c2 and take the reciprocal.
00:55
So we have 1 over 3 micro ferret plus 1 over 6 micro ferret and take the reciprocal.
01:03
We get 2 microferret.
01:08
And then for c34, we have 1 over c3 plus 1 over c4 and reciprocal.
01:19
So we have 1 over 2 microferret plus 1 over 4 microferret and we get 4 over 3 microferret.
01:30
So the effective capacitance, the equivalent capacitance will be c12 plus c34 because c12 and c34 are in parallel.
01:43
So we have 2 micro ferret plus 4 over 3 micro ferret.
01:51
Okay, we have 10 over 3 micro ferret.
01:54
Okay, so the total energy stored using u equals to half cb square.
02:02
Right, so u is half times 10 over 3 times 10 to pound negative 6 times 90 square and the answer is 0 .0135, three five juice.
02:17
Okay, so this is the answer for part a.
02:20
Next, i want to calculate the energy stored in each capacitor.
02:26
So we look at the circuit again.
02:37
So three micro ferret, six micro ferret, two micro ferret, four microferret.
02:45
Okay, so we know that the potential difference is 90 volts total...