Find c1 & c3 Given Vb, Vc, Vd, It, Xct Values

[harrypotter]

Homework Statement

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

Homework Equations

im not sure which equation to use

The Attempt at a Solution

im not sure what to do

i don't know if I am in the right forum

thanks

 

[berkeman]

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.

 

[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?

 

[berkeman]

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.

 

[harrypotter]

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

 

[Astronuc]

the Xc for c4 is 2.34k

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, X_c = 1/($\omega$C).

 

[harrypotter]

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, X_c = 1/($\omega$C).

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

 

[Astronuc]

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 X_t is the total reactance, and I_t 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 V_A, I_t and X_t?

 

[harrypotter]

...Va, It, Xt = Vs ?

 

[Astronuc]

...Va, It, Xt = Vs ?

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.

I_t*X_C6 is related to V_E.

Write some equations.

 

[harrypotter]

edit: nm