# Two capacitor problem

1. Sep 21, 2005

### bemigh

Hey guys,
I have this problem.. When two capacitors.. (different) are in parallel across a 10.0 volt rms, 1.0 khz oscillator, the currrent is 545 mA . When the same capactiors are in series.. its 126 mA. What are the values of the two capacitors...

Alright.. now... The formula when its in parallel is : I=wCV and C= C1+C2
When in series: its I=wCV and 1/C = 1/C1 + 1/C2

Using these formulas.. i plug in the known values or voltage, omega ( which i found) and the current... I have two equations with two unknowns... and after a lot of greasy quadratics.. i have two answers.. but it says they are both wrong.. 4..628e-6 and 3.522e-6

Did i stray anywhere here??
Cheers
Brentski

2. Sep 21, 2005

### CarlB

Brentski,

It seems to me that the capacitance values you got are way too small. They should add up to something like .0545 F, and should add in inverse to .0126 F. Maybe I'm off by a factor of 2 pi, but even then my values are WAY larger than yours.

1KHz is a very low frequency. 10V is a low voltage. But 126mA is a huge current. It's going to take a big capacitor for that kind of current.

Carl

3. Sep 22, 2005

Seems like there are some formulas missing? Are you looking to include capacitive reactance? 1 over the square root of 2 pi fc ? (or something like that) . In your equation, what does 'w' represent?

What happens if you plug in a range of arbitrary cap values? then plot the results and approximate whether to increase/decrease values? Id try but dont know what 'w' refers to

4. Sep 23, 2005

### Andrew Mason

Your expression for current is wrong. It should be: $I = V/X_c = V/\omega C$

So for the series configuration:

$$I_s = V(1/C_1 + 1/C_2)/\omega$$

and in parallel:

$$I_p = V/\omega(C_1 + C_2)$$

Plug in values for the two currents and V and work out C1 and C2

AM