# Calculate voltage across capacitors in series

#### matthewd

Joined Jun 19, 2015
2
Hello to All

My homework question is –

“A 3uF and a 6uF capacitor are connected in series to a 24 V a.c. 50Hz supply. Calculate the voltage across each capacitor”

I researched how to answer this one and found this –

The best answer states that -

Xc = 1/(2pi fc) = 1/[(6.28) x (20) x (.0001) = 79.6 ohm
Impedance angle = arc tan Xc/R = arc tan (79.6 ohm) / (56 ohm) = 54.87 degrees
Voltage across the capacitor = (sine of impedance angle) x (Input Voltage)
Vc = (sin 54.87 degrees) x 10 VRMS = 8.18 VRMS

I have done the first line of this equation and I have –

1/2xPiex50x0.000003= 1,061.03 Ohm

Now when I come to the second line of the solution I am totally stuck, I don’t have two impedance values to get the impedance angle from, so I don't really know how to proceed.

Or if anyone knows a simplier formula for calculating this one I would appreciate it - I have a feel I have missed something with my approach.

Thanks,
Matt.

#### crutschow

Joined Mar 14, 2008
32,939
Yes, you did miss something. You are looking for a solution before analyzing the problem and making it a much harder problem then it is. Note that, to solve an engineering problem, you don't always need all the information you are given.
Instead of looking for a canned solution, you need to think a little more about the problem.
It has only capacitors in the circuit so you don't need to do impedance or phasor calculations or even know the frequency of the power supply to solve that problem. You just have to know the relative AC impedance of the two capacitors, which is independent of frequency, and use that to calculate the relative voltage drop of each capacitor.
I trust you can do the rest.

Last edited:

#### WBahn

Joined Mar 31, 2012
29,183
You are trying to imitate a monkey imitating an engineer. Don't do that.

Look at the problem from basic principles and apply what you know, not what you can find that someone else claims to have done on a problem that is unrelated.

You are looking for formulas to throw numbers at without making any effort to even understand what the formula means, let alone whether or not it applies to your problem.

Note that one assumption that you do have to make is that the capacitors are initially uncharged (or that they at least have the same charge on them).

#### crutschow

Joined Mar 14, 2008
32,939
........................
Note that one assumption that you do have to make is that the capacitors are initially uncharged (or that they at least have the same charge on them).
But any initial charge doesn't affect the AC voltage division.

#### DickCappels

Joined Aug 21, 2008
10,068
With an ideal capacitor if one of both had initial charges the initial charge would remain as a DC offset to the voltage across the capacitor(s). Of course we can figure that the questioner was only referring to the AC component of the charge. Wait…since it is AC, can't we just say that the net change across each capacitor is zero?

#### WBahn

Joined Mar 31, 2012
29,183
If you have a voltage of 100 V DC and that voltage is going up and down sinusoidally by 1 V AC, what is the voltage on that capacitor?

#### crutschow

Joined Mar 14, 2008
32,939
I would assume the question is only referring to the AC voltage across the capacitors since no DC voltage is mentioned.

#### WBahn

Joined Mar 31, 2012
29,183
I would assume the question is only referring to the AC voltage across the capacitors since no DC voltage is mentioned.
Which is why I specifically stated that "one assumption you have to make is...."

#### shteii01

Joined Feb 19, 2010
4,644
Step 1.
Convert capacitances into impendances. I don't waste my time with stupid reactances.
Zc=1/(j*2*pi*f*C)

Step 2.
Use voltage divider formula.