# Critical thinking questions on Capacitance

Discussion in 'Homework Help' started by electronicstech07, Sep 28, 2008.

1. ### electronicstech07 Thread Starter Member

Nov 16, 2007
18
0
I am stuck on some critical thinking questions on Capacitance. I will work out the problem to see if I can get it right. Please do correct my mistakes along the way. I would appreciated it if someone can help me correcting my mistakes:

1. If Vs = 25V, Vc1= 10V, Vc2= 15V, CT=10 uF, determine each capacitance value in this series capacitive circuit.

I do know Q= V*C. So Q=25V * 10 uF = 250 uC. C=Q/V, C1 = 250uC/10V = 25uC, C2 = 250 uC/15V = 16.66 uC.

2. Two capacitors ( one 1 uF, the other unknown) are charged from a 12V source. The 1 uF cap is charged to 8V, the other to 4V. What is the value of the other capacitor?

In the chapter I was taught this formula Vx = Ct/Cx * Vs. Obviously this formula doesn't apply to this question. I know I can't find QT unless I know CT and VT, so I'm stumped on this question.

3. A 10 kHz voltage source is applied to a .0047 uF capacitor. 1 mA of rms current is measured. What is the value of the voltage?

I know V= I/XC. XC = 1/2*pi*f*C. So 1/6.28*10kHz*.0047uF = 3.38 kilo Ohm. V= 1 mA * 3.38 Kilo Ohm = 3.38 V.

4. A certain capacitor stores 50 millijoules of energy when charged to 12 V, what is the capacitance value?

I was taught this formula where Energy = 1/2*C*V^2. This question is one where I just don't know what steps to take to find the right answer.

Thanks again!

Last edited: Sep 29, 2008
2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
The Homework Help section asks that you post up the work you have done so we can see it and find where you went wrong, or suggest better approaches. We will not do all your work for you, no matter how little time is left before it must be submitted.

Thank you.

3. ### electronicstech07 Thread Starter Member

Nov 16, 2007
18
0
I understand the forum rules. I have gone back and tried my best to solve the questions and have edited my original post to include the worked out problems to the best of my knowledge up to this point in my studies.

Last edited: Sep 29, 2008
4. ### Ratch New Member

Mar 20, 2007
1,068
3
electronicstech07,

Capacitance is measured in farads not coulombs as you indicate above.

A capacitor is not charged, it is energized. That is, it can become imbued with energy. The same number of coulombs that are put on one plate are taken off the other plate for a net total of zero. The energy required to do this is stored in the electrostatic field which remains with the capacitor as long as an imbalance of charge exists between the plates. You know the same quantity of charge is displaced on both caps if they are in series. So what is the capacitance of the other capacitor if it measures 4V, or half the other's voltage? 2uF I would say.

V = I*Xc, not I/Xc . Otherwise you seem to have the gist of it.

50E-3 = (CV^2)/2 . I hope you can find C from that equation when V = 12 .

Ratch

5. ### electronicstech07 Thread Starter Member

Nov 16, 2007
18
0
Thanks for pointing that out. I did look at it again and have concluded it was a typo.

If the same amount of charge is displaced on both capacitors in a series circuit, then Q=V1 * C1 = 8V * 1 uF = 8 uC. Then C2 = Q/V2 = 8 uC/4V = 2 uF.

Another typo.

If the formula requires for C and V to be known, how can I manipulate the formula to find C? Very confused on this one. Thanks again!

Last edited: Sep 30, 2008
6. ### RimfireJim Member

Apr 7, 2008
22
2
By simple algebra, if it is even that (rearrange the equation to solve for C): Multiply both sides by 2, divide both sides by V^2, presto, you have C by itself on one side, everything else on the other.

7. ### electronicstech07 Thread Starter Member

Nov 16, 2007
18
0
Wow. That thought completely escaped my mind.

50 mJ = CV^2*0.5

= 50 mJ = C*144 * 0.5

=50mJ/72V = C

C = 694 uF

Thanks again!