I added a circuit that I drew in microsoft paint to add a visual. Sorry for its ugliness! ;D
V = I*Z
Ieff = I / sqrt(2)
Qreactive = V^2 / Z
Pavg = (power factor) * (Veff) * (Ieff)
Sapparent = ((Ieff)*(Veff))/2
The Attempt at a Solution
Since this is quite a long problem, I'll detail my thought process up until I get stuck. On the left, I have my work, and then I explain what I am doing on the right.
a) To find the effective current over the load:
P = (1/2)*v*i*(pf) Solve for i... (pf = power factor)
i = 2P/(v*pf) Then substitute in values...
i = (2*5000)/(300*0.6)
i =55.56 Now I have to divide the current by the square root of 2 to get the effective power
i = 55.56/sqrt(2)
i = 39.28 A rms This is the answer I get for a) and I have confirmed that it is correct.
b) Now I have to correct the power factor from 0.6 to 1 by adding a capacitor in parallel with the load.
PF = 1
theta = cos^-1(1) Finding the angle when the PF is 1...
theta = 0 degrees
Qn = tan(0deg) Using the new power factor, I'm finding the new reactive power
Qn = 0
Qc = -(Ql - Qn) We can find the reactive power of the capacitor by subtracting Ql and Qnew
Qc = -(6660-0) Substituting in values...
Qc = -6660 VAR This is the reactive power of the capacitor
Q = V^2 / Z Now we use the reactive power of the cap to find its impedance...
Z = V^2 / Q Solve for Z
Z = (300^2 / -6660) Substitute
Z = -j13.51 This is the impedance of the cap
Z = -j/wc Now we have to find the capacitance...
C = -j/wZ Solve for C
C = -j / (380 * -j13.51) Put the equation into a calculator...
C = 194.8 uF I have no way of confirming if this is right, but I'm pretty confident that it is!
c) Ok... this is where things start to get a bit shaky...
So in order to find the effective current of the capacitor...
V = I * Z
I = V / Z Solve for I...
I = (300) / (-j13.51) Substitute...
I = j 22.2058 Then I have to multiply this by 1/sqrt(2) to get the effective current...
I = j 22.2058 / sqrt(2)
Ieff = 15.7 A This is the answer I get; however, I know it is wrong. It should be twice that...
Ieff = 31.4 A This is what the answer is supposed to be. I am unsure why I get half it.
d) I feel as though the snowball is growing bigger and bigger as it rolls down the hill... Or in other words, I believe a small mistake that I've made is making all the following questions incorrect.
In order to find the impedance of the load...
V = I * Z
Z = V / I Solve for Z
Z = (300) / (39.28) Substitute... (But I don't know which value to use for I... so I used Ieff)
Z = 7.64 ohms This is the impedance of the load which is in parallel with the capacitor...
Ztotal = (7.64) || (-j13.51) Finding the total impedance...
Ztotal = 5.79 - j3.27 Pretty sure this is wrong...
e) So now I need to find the effective current of the source...
V = I * Z
I = V / Z Solve for Z
I = (300) / (5.79 - j3.27) Substitute...
I = 39.28 + J22.19 Now I have to multiply this by 1/sqrt(2) to get the effective power...
I = 27.78 + j15.69
I = 27.78 A rms I know this is wrong. The answer should be 23.57 A rms.
f) Now I need to find the apparent power, S, for each the load, capacitor, and the source...
S = (Ieff * Veff) / 2
Scap = (300*31.4)/2
Scap = 5.71 kVA
Sload = (300*39.28)/2
Sload = 5.892 kVA
Ssource = (300*23.57)/2
Ssource = 3.535 kVA I know this value is wrong since it is supposed to be 5kVA. And because this is wrong, I suspect that the others are wrong as well.
g) And the last one... It's more of an explanation, so I'll take a whack at it...
Ssource should not equal Scap + Sload because capcaitors only absorb reactive power (Q) and the Ssource is not a complex number. However, I'm conflicted on this reasoning because isn't it also the case that the power delivered should be equal to the power absorbed?
OK... So it is a pretty long problem and I appreciate it if you have stuck through the whole thing. I don't know exactly what is wrong, but I have suspicions:
1. I am unsure as to why the answer I got is exactly one half of the actual answer... perhaps I dropping a 2 somewhere? I know there are a whole bunch of new equations that get thrown around in AC Power, so I may have gotten a formula wrong.
2. For part c), I really don't think I am supposed to substitute the I effective in for I, but I don't know what I would be.
3. And then everything from part d) and on I think just stems from the previous wrong answers.
Again, thank you to all those who have been helping me through circuits! It's almost over! (^_^)