Series Heating Effect vs. PF...

Oh come now.

"It's a common question, and the stream of conflicting answers that is sure to follow will only add to the confusion!"

"220 VAC Electric Baseboard Heater Without A Ground" seemed like a simple yes or no question too.

We are up to 112 "answers" and counting on that one.

Nothing's ever simple.

Agreed, but with one caveat; i.e., "Not on CR4" where opinion trumps over-rides fact more often than not! I'm sure that you get an occasional chuckle watching how "engineering facts" can get so easily confused with "engineering opinion".

Chuckle and guffaw!

Hey, we all have our moments.

Lyn, that was a " fun" question.

Did you see how many in the group came out to play :)

All that leading and lagging theory sounds very old school for us, power electronics guys. Since nowadays there are plenty of modulated Power Supplies, the definition of PF changed a long time ago.

That being said, your argument is true taking into account the frequency response of the element in question.

The power factor won't come into play for the series resistance, but can the other choice be correct, either?

Is the first answer the only other option to the second answer?

Does nominal rating usually/ever determine actual power/current/losses/heat?

If you have a power consuming item in a circuit consuming close to its 180 watt nominal rating and then you add a nonnegligible series resistance what conditions would be needed to keep the power consuming item consuming the same power? Wouldn't that have an effect on the heating of the series resistance and doesn't that depend not solely on the power consumption of the power consuming items, but in large part on the size of the series resistance added?

Looks like a trick question.

What is constant is the applied voltage. Power equals voltage squared divided by impedance. Power will have a real and imaginary part. If your heater has resistance plus inductance then the impedance will be R + jX. Let's set the voltage to 1 to make calculations simpler.

P = 1 / (R + jX).

multiply by (R-jX)/(R-jX) to make the denominator real

P = (R-jX)/(R²+X²)

the real part (which generates heat) is R/(R²+X²)

Adding series resistance increases R. As R becomes larger, the denominator increases faster than the numerator, so the total real power becomes less.

OK, thought about it some more and ran some numbers. Here is a simpler explanation. The same current goes through the heater and added resistance. So the power (current x voltage) is proportional to the impedance V = I (R + jX).

Total power = 400 VA and real power = 180 W for heater alone.

Power = 180*R + 357j *X [ sqrt(180²+357²)=400.]

The total power and real power will be proportional to the resistance R and inductive reactance X.

It turns out that as you add series resistance, the total power increases to a point and then decreases. Using the formula for power:

P = (R-jX)/(R²+X²) and plotting the real power

I've been trying to find a simple answer.

If voltage is constant, power is inversely proportional to Impedance: P = V²/Z

If impedance is real (resistance only), if resistance increases, power decreases.

If impedance is complex (resistance + reactance), then if resistance increases (real part of impedance) then it does not mean that real power increases. Here's an example (Complex plot of Z in blue and Power in Red)

You're way overthinking this, the OP's question was whether the current from the real power, or the current from the apparent power, causes the the resistor to heat up.

The answer is the current from the apparent power (S). Surely you don't think that if the load were purely inductive (or capacitive for that matter) then there wouldn't be any heating.

If you start with a pure inductor and add resistance, the total power dissipation (real power) will rise to a maximum when the resistance equals the inductive reactance. After that point, it will decline.

The OP had a heater which had resistance and inductive reactance, and in his case, the inductive reactance was greater than the resistance. (400 VA and 180W equates to 180 real power and 353 reactive power.) So adding more resistance caused the total (real) power to increase.

If there had been more resistance than inductive reactance in the heater, adding external series resistance would have resulted in a decrease in real power as it would have been on the downward slope of the power curve.

Impedance and Power on the complex plane:

Impedance = 100j + Resistance

Resistance = 0 - 500

Real Power as a function of Resistance:

Wait, if the load is purely inductive (or purely capacitive), so no resistance, what mechanism is responsible for the heating?

Power is a complex quantity if inductance or capacitance are present in the circuit. You are correct. If the load is pure inductance, power is purely imaginary and there is no heat. (See first plot in #12, at zero resistance power is 0 - 400j)

"...power is purely imaginary and there is no heat..." in the inductor, of course, but real power (Watts) are dissipated in the series resistor that has the reactive current passing through it to reach the inductor, regardless of the angle of the current relative to the voltage.

If we continue with the philosophy around the power losses in the inserted resistance we have to take the power factor into consideration.

If you intend to keep Srat and Prat you have to keep V2=V2rat [and I=Irat] and then you have

to rise the supply voltage V1.

In this case since the I=Irat [constant] the power factor does not matter indeed.

However, if you cannot do it, that means the supply voltage V1=Vrated the voltage V2 will be

less and the current will be less [and so S,P and Q] and then the losses in R1 will be less.

If Q=SQRT(S^2-P^2)=SQRT(400^2-180^2)=357.2 VAr

P=V2^2*R2/SQRT(R2^2+X2^2)

Q=-V2^2*X2/SQRT(R2^2+X2^2)

Q/P=X2/R2

P=V2^2/R2/SQRT(1+(Q/P)^2) R2=V2^2/P/SQRT(1+(Q/P)^2)

X2=R2*Q/P

V1=SQRT((V2+R1*I*COS(Fi2))^2+(R1*I*SIN(Fi2))^2) if you know V2

V2=sqrt(V1^2-(R*I*SIN(Fi2))^2)-R*I*COS(Fi2) if you know V1

Nevertheless V2<V2rat R2 and X2 remain the same[usually].

In this case I=V1/SQRT((R1+R2)^2+X2^2)

S=I*V2 P=S*cos(Fi2)

Dude, you need to wake up!!!

See the wattage of your resistors. Simple series circuitry will tell you half the voltage, Amperage goes twice. That's the source of your thingy heat up.

Example, rewire all your house wiring and make it in series. What do you think will happen to your appliances?