Confusion in relation between Frequency, current and Power factor relation

[gauthamtechie]

First let me state my inferences on these:

If the frequency increases, the reactive component increases. So Power factor = Resistance/Impedance decreases. Thus Power factor decreases -> which means the current drawn is more, considering load is inductive in nature

But we know I = V/ 2*pi*f*L . So if frequency increases, the current drawn decreases as per this relation.

What am I missing here?

 

[crutschow]

If the inductor is in series with resistor then, as the frequency increases, the inductive reactance (impedance) increases so the current draw is less and so is the resistive power. The power factor also decreases since the inductive reactance is a larger percentage of the total series impedance.

If the inductor is in parallel with the load, then the current draw is also less, but now the power factor increases since the resistive current remains the same but the inductive current is a smaller percentage of the total current.

The current draw is only more with a lower power factor if the resistive power remains the same and that is not the case for either of the above scenarios.

 

[gauthamtechie]

The first scenario I understand.

The second scenario:

If the inductor is in parallel with the load, then the current draw is also less, but now the power factor increases since the resistive current remains the same but the inductive current is a smaller percentage of the total current.

It makes perfect sense that the inductive current is smaller, hence Power factor increases. But again, another way of expressing Power factor is ratio of the Resistance(R) to Total Impedance(Z). Here with a larger reactance in parallel, the Total impedance increases as Frequency increases. So Power factor as R/Z must decrease as per this relation. Again I think I'm missing something.

The current draw is only more with a lower power factor if the resistive power remains the same and that is not the case for either of the above scenarios.

In the second scenario, with a parallel resistance, doesn't the resistive Power remain the same?

 

[MrAl]

First let me state my inferences on these:

If the frequency increases, the reactive component increases. So Power factor = Resistance/Impedance decreases. Thus Power factor decreases -> which means the current drawn is more, considering load is inductive in nature

But we know I = V/ 2*pi*f*L . So if frequency increases, the current drawn decreases as per this relation.

What am I missing here?

Hello there,



As Carl was saying, the relationships here depend on how the circuit is arranged, the topology of the circuit. We can look at two cases and see what happens. The simplest two cases are the series RL and the parallel RL so we'll do that.

First, the impedance of the parallel circuit is:
Zp=(j*w*L*R)/(R+j*w*L)

and the impedance of the series circuit is:
Zs=R+j*w*L

The phase shift for the parallel circuit is:
THp=-atan(R/(w*L))

and the phase shift for the series circuit is:
THs=-atan(w*L/R)

Since the power factor is cos(TH) where TH is the phase shift between current and voltage, the power factor for the series circuit is:
PFs=R/(sqrt(R^2+w^2*L^2)

while for the parallel circuit we have a power factor:
PFp=w*L/sqrt(R^2+w^2*L^2)

Now we are in a position to evaluate the power factor directly in either the series or parallel case.

Starting at zero frequency, for the series case we get:
PFs(0)=R/R=1

and at infinite frequency we get:
PFs(inf)=0

and for the parallel case at zero frequency we get:
PFp(0)=0

and for infinite frequency we get:
PFp(inf)=1

Now we can compare the two sets of information.

We immediately see that the power factor for zero and infinite frequencies are reversed for the series and parallel RL circuits.

Now we look at the current...

The current in the series circuit is:
Is=1/sqrt(R^2+w^2*L^2)

and in the parallel circuit it is:
Ip=sqrt(R^2+w^2*L^2)/(w*L*R)

The current in the parallel circuit for zero and infinite frequencies is:
Ip(0)=infinite
Ip(inf)=1/R

The current in the series circuit for zero and infinite frequencies is:
Is(0)=1/R
Is(inf)=0

So you can now compare these limits and see that the responses are different.

Listing the values all on one line for parallel in order of PF for zero and inf freq, then current for zero and inf frequencies, followed by the series case on the next line:
0,1,inf,1/R (parallel)
1,0,1/R,0 (series)

We can see that the results are sort of swapped except for the current which is either infinite or zero, but the 1/R currents are also swapped. So the two circuits have a basic difference in that both their power factors AND current responses are swapped.

 

[Ratchit]

gauthamtechie,

But again, another way of expressing Power factor is ratio of the Resistance(R) to Total Impedance(Z).

No, it is the Arccos(R/Z).

Again I think I'm missing something.

Correct, you are indeed. PF is determined by voltages across the components in a series circuit and current through the components in a parallel circuit. Voltage is proportional to impedance in a series circuit and current is proportional to admittance in a parallel circuit. Therefore, you have to convert impedances to admittances in order to compute PF in a parallel circuit.

Ratch

 

[crutschow]

........................................
In the second scenario, with a parallel resistance, doesn't the resistive Power remain the same?

True. Don't know why I stated that.