# Transformers and Active/Reactive power question from HW

1. Jan 28, 2017

### JoeMarsh2017

1. The problem statement, all variables and given/known data
A 7200-240 V, 60 Hz transformer is connected for step up operation, and a 144∠46° Ω Load is connected to the secondary. Assume the transformer is ideal and the input voltage is 220 V at 60 Hz.
Determine
a)secondary voltage
b)secondary current
c)primary current
d)input impedance
e)1)active
2)reactive
3)apparent power input to the transformer

2. Relevant equations
since its operating as "Step Up" transformer...the turns ratio aka "a' = 30

240-7200
1:30 step up

3. The attempt at a solution
a)secondary voltage =220*30= 6600 V∠0°
b)secondary current = 6000V/144Ω ∠46°= 45.8333 A∠-46°
c)primary current= a x I sec = 45.83333 * 30 = 1375 A∠-46°
d)Z input= 220 V / 1375A = 0.16∠46°Ω

Active / reactive and apparent Power
are all listed as answers in the back of the book

210.1 kW =active power is my true power
217.6 kVAR = Is my reactive power
302.5 kVA 1375*220 = 302500 /1000(for the K) = 302.5 kVA

Can you kindly show me what I am forgetting about active and reactive power guys?

Thanks Joe

2. Jan 28, 2017

### goodphy

One-cycle-averaged power is conveniently calculated by using complex representations of voltage and current such as $$\tilde S = \frac{{\tilde V{{\tilde I}^*}}}{2} = P + iQ$$, where P and Q are an active and reactive power respectively, and absolute value of S is an apparent power.

Here, I choose that V and I are for the secondary side of the transformer so, $$\tilde V = 6600{e^{i0}} = 6600,{\rm{ }}\tilde I = 45.8333{e^{i\left( { - {{46}^ \circ }} \right)}}$$

Calculation shows that P ≅ 105 kW, Q ≅ 109 kW, and the apparent power is 151 kW.

Your values for powers are twice than what I calculated. I guess you missed a factor of 1/2 somehow.

3. Jan 28, 2017

### JoeMarsh2017

210.1 kW
217.6 kVAR

These are the answers from the book

4. Jan 29, 2017

### magoo

Since you are dealing with effective or rms voltages, the power equation is:

S = V * I' = P + j Q

I' should be I-conjugate

There is no factor of 2 involved.

5. Jan 30, 2017

### JoeMarsh2017

GUYS, I figured it out...My problem was that once I solved for "S"..and I had the angle @ 46°...then I can solve for active and reactive power
Power triangle means
s=302.5 theta=46°
Active=210.13 =302.5cos 46°
Reactive=217.34 =210.13tan 46

6. Jan 30, 2017

### cnh1995

7. Jan 30, 2017

### JoeMarsh2017

210.13 tan 46 gets you 217.59 which is what I was looking for..

210.1 kW Active Power
217.6 kVAR Reactive Power

This is per the power triangle

8. Jan 30, 2017

### cnh1995

Oh..Right!
You took Q=PtanΦ and I misread it as Q=StanΦ.
Q=S*sinΦ=P*tanΦ..