# power factor, voltage and current waveforms

Discussion in 'Homework Help' started by m_p, Mar 5, 2011.

1. ### m_p Thread Starter New Member

Nov 23, 2010
6
0
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

2. ### Vahe Member

Mar 3, 2011
75
9
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

Last edited: Mar 5, 2011
3. ### Vahe Member

Mar 3, 2011
75
9
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

4. ### m_p Thread Starter New Member

Nov 23, 2010
6
0
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

5. ### Vahe Member

Mar 3, 2011
75
9
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

m_p likes this.
6. ### m_p Thread Starter New Member

Nov 23, 2010
6
0
thanks Vahe, i understand it now.