AC Circuit Voltage Waveform Analysis

[james123]

Homework Statement

An A.C. voltage, V, comprises a fundamental voltage of 100V rms at a frequency of 120 Hz, a 3rd harmonic which is 20% of the fundamental, a 5th harmonic which is 10% of the fundamental and at a phase angle of 1.2 radians lagging.

1. (i) Write down an expression for the voltage waveform.
   (ii) Determine the voltage at 20 ms.
   (iii) Given an ideal V = 100 V rms, what is the percentage error at 20 ms?
   

Homework Equations

Vm=Vrms*√2

V=Vm*Sin(ωt) volts

The Attempt at a Solution

(i)[/B] Vm=100*√2=141.4 volts at 120Hz

3rd Harmonic= 20% of 141.4= 141.4*0.02= 28.3 volts at 360 Hz

5th Harmonic= 10% of 141.4= 141.4*0.10= 14.1 volts at 600 HzSo, V= (141.4*Sin(240πt)) + (28.3*Sin(720πt)) + (14.1*Sin(1200πt)) ?(ii) Fundamental at 20 ms= 141.4*Sin(240πt)= 83.11 volts

3rd Harmonic at 20 ms= 28.3*Sin(720πt)= 26.91 volts

5th Harmonic at 20 ms= 14.1*Sin(1200πt)= -13.14 voltsSo, V at 20 ms= 83.11+26.91-13.14= 96.9 volts ?

(iii)

(96.9-83.11)/83.11*100= 16.59

So, percentage error at 20 ms= 16.6% ?If anyone could confirm my answers/tell me where I've gone wrong, I'd greatly appreciate it!

Many thanks in advance!

 

[gneill]

You haven't accounted for the phase angle associated with the 5th harmonic. Otherwise your work looks okay.

 

[james123]

Thanks for replying! Ahh I see.

So,
(i) 5th Harmonic= 10% of 141.4= 141.4*0.10-1.2= 12.9 volts at 600 Hz

Therefore V= (141.4*Sin(240πt)) + (28.3*Sin(720πt)) + (12.9*Sin(1200πt-1.2)) ?

(ii) 5th Harmonic at 20 ms= 12.9*Sin(1200πt-1.2)= -12.02volts

Therefore V at 20 ms= 83.11+26.91-12.02= 98volts ?

(iii) (98-83.11)/83.11*100= 17.91

So, percentage error at 20 ms= 17.9% ?Does this look better gneill?

 

[gneill]

I don't understand your use of 1.2 in your first step. 1.2 is an angle in radians while the other term is a voltage. So your value of 12.9 V for the magnitude of the 5th harmonic is not correct. The phase angle should not affect the magnitude.

 

[james123]

Oh right I see, so will that remain as it was originally?

How did the other calculations look to you?

 

[james123]

I think I see what you're saying. That the answers are correct I just haven't included the -1.2 in the calculations on here?

My mistake if that is the case?

 

[gneill]

Oh right I see, so will that remain as it was originally?

Yes. The phase is applied when you calculate the angle argument for the trig function (sin() in this case). The magnitude of the term is not affected by the phase.

How did the other calculations look to you?

Fine.

I think that if I were doing the problem I'd probably hold off applying the $\sqrt{2}$ to the individual terms so that I wouldn't have to carry along so many decimals through the work. And I'd set a variable $ω$ as the fundamental angular frequency. So something like:

$ω = 2 \pi⋅120~rad/sec = 240 \pi~rad/sec~~~~~~~φ = -1.2~rad$

$V(t) = 100~\sqrt{2}~V \left( sin(ω t) + \frac{20}{100} sin(3 ω t) + \frac{10}{100} sin(5 ω t + φ) \right)$

Then I would have the option of dropping the $100~\sqrt{2}~V$ term when calculating the percent error, since it will cancel out as it occurs in both the numerator and denominator of that calculation.

I think I see what you're saying. That the answers are correct I just haven't included the -1.2 in the calculations on here?

My mistake if that is the case?

You just have to apply the phase shift appropriately, namely in the calculation of the 5th harmonic's contribution at the specified time. Otherwise you are on the right track.

 

[james123]

Ahh I see what you mean now. Makes sense.

Unfortunately with my course it's their way or no way, so more often than not we find ourselves doing things long winded!

Thanks for all your help gneill!