AC power - Simulation and hand calculated values different

[eehelp150]

Yes, but I imagine that the algebra could get tedious. You want to find the reactive power as a function of C, so that means finding the imaginary part of the complex power. It would be easier to plot the reactive power and pick out the solution on the plot.

I am plugging in the original total real power and total reactive power. I cannot get the original value of capacitance.

Edit:
I used 576-430.8i instead and got the original capacitance value. How do I solve for C if I don't have a "defined" real power to solve for? Do I simply divide that by 2 (50%) as well?

 

[gneill]

I used 576-430.8i instead and got the original capacitance value. How do I solve for C if I don't have a "defined" real power to solve for? Do I simply divide that by 2 (50%) as well?

No, the real and imaginary parts of the power will have some complicated relationship.

Since you want to reduce the reactive power by 50% you can simplify things a bit by setting Vs = 1. Then ignoring units the current will be $I = 1/Z$ and the power will be $P = V_s I^* = I^*$.

Isolating the imaginary part of $I^*$ might be tedious. You can write your impedance as:

$Z = 12 + j20 + \frac{1}{\frac{1}{20} + jωC}$

Might as well define $x = ωC$, then
$$I = \frac{1}{12 + j20 + \frac{1}{\frac{1}{20} + jx}}$$
Normalizing that to pull out the complex component is not complicated, just labor intensive. Then find its value using the original capacitor value in x = ωC as the "benchmark" value for the imaginary component. You'll want to find a new value for x that yields half of that power.