Finding missing caps

1. Oct 12, 2007

harrypotter

1. The problem statement, all variables and given/known data

find c1 & c3

known values:

c2= .01uF
Xc= 2.34k
c5= .047uF
c6= .033uF

given these values:

Vb 21.91v

Vc 17.51v

Vd 14.48v

It 1.76mA

Xct 28.34k ohms

http://i216.photobucket.com/albums/cc193/harrypotter33/Untitled-1.jpg [Broken]

2. Relevant equations

im not sure which equation to use

3. The attempt at a solution

im not sure what to do

i dont know if im in the right forum

thanks

Last edited by a moderator: May 3, 2017
2. Oct 12, 2007

Staff: Mentor

Capacitive reactance varies with frequency, so you need to know what frequency is being used by the AC source. Otherwise, you cannot mix Xc impedances and C capacitances in the same capacitive voltage divider equation. Do you have some way to determine the frequency? Like, are you given both Xc and C for any of the caps (I don't see that)?

Welcome to the PF, BTW. And yes, you picked the right subforum for your question.

3. Oct 12, 2007

harrypotter

thank you for the reply. the Xc for c4 is 2.34k. so, i need to figure out the frequency first? but how do i figure this out?

4. Oct 12, 2007

Staff: Mentor

I'm not tracking the whole problem very well, but it looks like they have given you a mixture of capacitances and Xc impedance values, and are asking you to calculate two unknown capacitance values. If you had all the known capacitors as capacitance values, then the frequency of excitation would not matter, and you could just calculate the unknown cap values using the relative impedances. But if you are only given information about a couple of the capacitor positions as Xc values, then you need to know the frequency of excitation that the Xc is given for. You would use the frequency to get back to the capacitor values for those positions.

5. Oct 13, 2007

harrypotter

this is more confusing than trying to remember the logic gates. anyone want to try and solve this for me please? :)

6. Oct 13, 2007

Astronuc

Staff Emeritus
but what is the capacitance of c4?

In the OP one states c4 = 2.34k. Are the units $\mu$F, which would be an impressive capacitor, or pF?

Capacitive reactance, Xc = 1/($\omega$C).

7. Oct 13, 2007

harrypotter

oh sorry, the capacitance of c4 wasnt given. but was given an Xc of 2.34k instead

8. Oct 13, 2007

Astronuc

Staff Emeritus
Well there are two loops and two unknowns, C1 and C3.

Take the potenials with respect to ground below the voltage source.

I take it the Xt is the total reactance, and It is the total current, which must pass through the voltage and those capacitors which are in series with the voltage source.

What can one say about VA, It and Xt?

9. Oct 13, 2007

harrypotter

...Va, It, Xt = Vs ?

10. Oct 13, 2007

Astronuc

Staff Emeritus
Well Va = Vs, assuming the ground reference is at the bottom left of the figure, i.e. inlet to source.

It must go through C1, C5 and C6.

It*XC6 is related to VE.

Write some equations.

11. Oct 15, 2007

harrypotter

edit: nm

Last edited: Oct 15, 2007