hello,
i need to draw the waveforms of the voltage and current in an inductive circuit which has a power factro of 0.866 (which give a phase angle of 30 degrees).

i know that the current lag 90 degree behind the voltage, but i am not quite sure i understand how the power factor affect the waveforms.
any help would be much appreciated.
thanks
mario

 

I think you should consider a series LR circuit. Apply a sinusoidal input to the series circuit in the form of
\( v_{in}(t) = V_m \sin ( \omega t) \).
Then compute the current that is being drawn by the LR circuit from the voltage source. The input current should have the form
\( i_{in}(t) = I_m \sin ( \omega t - \theta) \).
In other words, the current will lag the voltage which is what you expect in an inductive circuit. Power factor is simply \( \cos \theta \).

Cheers,
Vahe

 

If you compute the power factor of the series LR circuit, you should get
\(
PF = \frac{R}{\sqrt{(\omega L )^2 + R^2}}
\)
where \( \omega = 2 \pi f \) and \( f \) is the frequency in Hz.

In your case you are given the power factor, so you have to pick a frequency and a value for R and you can solve for the L that can give you the desired power factor. You may also be given the frequency (for example, 60 Hz) in that case, you will still have 1 equation and 2 unknown. So you choose the R and compute the L value, for example.
There is no unique solution unless one of the elements are specified and a
frequency is given.

Cheers,
Vahe

 

hi vahe, thanks for your reply.
i think my tutor wants us to just draw a standard plot of voltage/current like the one below, but it also gave us the value of the power factor... i don't know why but i assume it must be shown somewhere on the graph

 

Given that graph, first mark the time difference between the plots. Call this time difference \( t_d \). Note that this time difference will be some fraction of the total period of the waveforms for the inductor current and voltage. If the waveforms have frequency \( f \), then the period in seconds will be \( T = 1/f \). Now we note that the full period \( T \) corresponds to \( 2 \pi \) radians or 360°. Therefore \( t_d \) corresponds to \( 2 \pi t_d / T\) radians or \( (t_d/T)\)360°. This is exactly the angle \( \theta \) and the cosine of this angle is the power factor.

If you do this for the waveforms you show, you will find that \( td = 0.25 T\). In other words, the peaks have a difference of a quarter of a period. So the angle theta is one quarter of 360° or just 90°. The power factor here is 0.

In your case you are asked to provide something that has power factor of 0.866; therefore, \( \cos \theta = 0.866\). This means that \( \theta = \cos^{-1} 0.866 = \pi/6\) radians or 30°. This means that your peak to peak difference has to be 30°/360° or 1/12 of the period. The actual time difference will depend on the frequency you choose but 30° will always correspond to the 1/12 of the period.

Cheers,
Vahe

 

thanks Vahe, i understand it now.