Series Parallel Capacitors

thatoneguy

Joined Feb 19, 2009
6,359
Is the voltage across C3 5V? If so, then it's correct. If not, then it's incorrect, because you can't end up with more voltage than you started with, the voltages need to sum to zero/source voltage.
 

praondevou

Joined Jul 9, 2011
2,942
Where do the 2.5mA come from?
Why not go the easier way?


Xc4 = 1/2pi x f x C4

Vc4 = Xct4 x Itotal

Vc3 = Vtotal - Vc4

This is getting confusing. Don't you want to post the X of every capacitor and we start from there?
 

thatoneguy

Joined Feb 19, 2009
6,359
This is getting confusing. Don't you want to post the X of every capacitor and we start from there?
I did that in post 14, the impedance is correct, I forgot an inversion for the total capacitance.

The total current of 4.188 mA and total impedance of 2.388k is correct, which are the answers he got ±1 in last significant digit. So he has it right, I'm not sure if he understands the full concept (that it can be worked by calculating capacitances, or impedances, then figuring out the rest)

The rest should be easy to figure out.
 
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thatoneguy

Joined Feb 19, 2009
6,359
XC1 is 1592 ohms, XC2 is 1062ohms, XC3 is 3980 ohms and XC4 is 796 ohms
Correct. and you ended up with the correct total current.

The total current will be going through C4/Z4, dropping some voltage.

The rest of the voltage will be across (C1/Z1 + C2/Z2) || C4, with total current split between those two legs.
 

Thread Starter

paul_alan

Joined Nov 5, 2011
43
Okay...Ct - 1.333μF, XCt - 2.389kΩ, IC4 (It) - 4.186mA, VC3 - 6.67V, VC4 - 3.33V.
taking the 6.67V across C3 and using it for B1 (C1, C2) then divide it over XC1,2. So,
6.67V / 2639 ohms equals 2.527mA. C3 - 6.67V / 3980 ohms equals 1.675mA. and those two currents totaled equal the current total of 4.18mA....i think i've got it
 
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