## 3rd and 5th harmonics

1. The problem statement, all variables and given/known data

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

1. Write down an expression for the voltage waveform.

2. Determine the voltage at 20ms.

2. Relevant equations

3. The attempt at a solution

Im not sure if this is right and would appreciate if someone could take a quick look over it please.

1. I have the fundamental (Vrms x √2) at 141.4v at 120Hz
making 3rd harmonic (20% of 141.4) = 28.3v at 360Hz
and the 5th harmonic (10% of 141.4v) = 14.1v at 600Hz

Therefore V=141.1 sin (2∏ft) + 28.3 sin (2∏ft) + 14.1 sin (2∏ft-1.2)

2. Using the equation above and inserting 20ms for "t" I got the following voltages

4.05v + -1.16v(?) + 74.15v = 77.04v

Any help would be appreciated thanks

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 Quote by jitznerd Therefore V=141.1 sin (2∏ft) + 28.3 sin (2∏ft) + 14.1 sin (2∏ft-1.2)
You're pretty close. But all of your terms are at the fundamental frequency f. How can you modify the 2nd and 3rd terms, to be at the 3rd and 5th harmonic frequencies instead?
 2. Using the equation above and inserting 20ms for "t" I got the following voltages 4.05v + -1.16v(?) + 74.15v = 77.04v Any help would be appreciated thanks
For the first term, I get something different. Can you show more explicitly how you calculated that one term?
I didn't calculate the 2nd and 3rd terms yet, since your original expression was incorrect. However, the 3rd term has an amplitude of just 14.1V, so no way can you get a contribution of 74.15V from it

Hope that helps.

 Thanks for that So would it be along the lines of: V=141.1sin(240πt)+28.3sin(720πt)+14.1sin(1200πt-1.2) How im working out the voltages is using above formula, This is how I worked out first section but pretty sure im going around this the wrong way 141.1sin = .268 240 x π x 0.02 = 15.08 .268 x 15.08 = 4.04v

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## 3rd and 5th harmonics

 Quote by jitznerd Thanks for that So would it be along the lines of: V=141.1sin(240πt)+28.3sin(720πt)+14.1sin(1200πt-1.2)
Looks good!

 How im working out the voltages is using above formula, This is how I worked out first section but pretty sure im going around this the wrong way 141.1sin = .268 240 x π x 0.02 = 15.08 .268 x 15.08 = 4.04v
To evaluate 141.1sin(240πt):
1. Evaluate (240πt) = 240 x π x 0.02 = 15.08, as you did.
2. Take the sine of that result, 15.08, but first be sure that your calculator is in radians (not degrees) mode.
3. Multiply the result of Step 2 by 141.1