- #1

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## Homework Statement

I added a circuit that I drew in microsoft paint to add a visual. Sorry for its ugliness! ;D

## Homework Equations

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! (^_^)