# power factor, voltage and current waveforms

#### m_p

Joined Nov 23, 2010
6
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

#### Vahe

Joined Mar 3, 2011
75
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:

#### Vahe

Joined Mar 3, 2011
75
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

#### m_p

Joined Nov 23, 2010
6
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 #### Vahe

Joined Mar 3, 2011
75
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