Curious3141 said:The slow way to do it is from first principles.
The quick way is to use the AC impedance of a capacitor, which is ##\displaystyle \frac{1}{j\omega C}##, which can be treated like a resistance, then apply the delta-star conversion formula to it and see what you end up with. Remember to convert the final impedance you get back to a capacitance by reversing the formula.
Looks like it's just a bit of algebra. Logically, you know that ω should ultimately play no role in the capacitance values you obtain, so it should cancel out along the way. Might as well just eliminate it from the start (or choose a convenient working frequency such as ω = 1). That may make things less cluttered looking.Greg Arnald said:there for if capacitors.. 1/ωCa=(1/ωC2*1/ωC3)/(1/ωC1+1/ωC2+1/ωC3). (??)
How do I convert this back to capacitance so I just have Ca=...?
The formula for delta star in capacitors is: Cdelta = C1 + C2 + C3, where C1, C2, and C3 are the capacitance values of the three capacitors in the delta configuration.
The delta star formula is different because it takes into account the connection of three capacitors in a triangular shape, while the series and parallel formulas only consider two capacitors connected in a line or side by side.
Yes, the delta star formula can be extended to any number of capacitors in a delta configuration. The general formula is: Cdelta = C1 + C2 + ... + Cn, where C1, C2, ..., Cn are the capacitance values of each capacitor.
The advantages of using a delta configuration for capacitors include: a higher capacitance value compared to a single capacitor, better stability, and the ability to handle higher voltages.
Yes, the delta star formula can be used for capacitors with different values. However, the resulting capacitance will be different from the individual capacitance values, and it may be necessary to use a conversion factor to accurately calculate the total capacitance.