Blow dryers and other elementary things

1. Jun 11, 2005

cepheid

Staff Emeritus
I can't seem to apply what I've learned in EE to the most elementary problems. I have two questions.

The first is based on an observation. When we plug in our blow dryer, and turn it on, the lights dim noticeably. Why?

My attempt to think about it involved connecting another load in parallel with the lamps and seeing what happened to the current going through each of the lamps, assuming (in my model) that there were only two of them, and they were identical. If the current through each lamp was reduced, I would have my explanation. But how to model the blow dryer? I had no idea. But I noticed the ratings written on it: 1600 W, 125 VAC, 60 Hz (there was also an option to set it for 250 VAC and 50 Hz). Why 125 V? Isn't 120 V standard for N. America? No matter...in my model, to simplify things, I used unrealistic numbers. But I digress. Back to the problem of how to model the blow dryer. Since 1600 W was such a large number, and the blow dryer was visibly affecting the electricity available, I decided that it must be drawing a lot of current. So I modelled it as a resistor with very low resistance, compared to the two lamps. So here are my unrealistic numbers I used. Source: 10 V
2 Lamps: 10 ohms each. Blow dryer: 0.1 ohms.

current through each branch is the conductance, G_i, times the voltage, V_i, across the ith branch. V_i = V = 10 volts for all i, so the currents are:

I_1 = (10 V)(1/10) = 1 A

I_2 = I_1 = 1 A

I_3 = 10 X 10 = 100 A

By KCL, I_total = I_1 + I_2 + I_3

V X G_total = (V X G_1) + (V x G_2) + (V X G_3)

= V X (G_1 + G_2 + G_3) = 102 A, which was obvious from the very first line of equations, but I elaborated to show that conductances in parallel add together, since we're adding more conducting pathways to the circuit, and this is equivalent to saying that for resistances in parallel, the overall resistance is lowered i.e. instead of G's, it could have been written: 1/R total = 1/R1 + 1/R2 + 1/R3. We all know this.

The problem I'm having is that the current through each lamp is the same as it would be if the blow dryer were not connected in parallel with them, and it's never going to change. Even if the third load you connect in parallel with the two lamps has an even smaller resistance than 0.1 ohms, that will only change the *overall* current drawn by the circuit. Each branch still "sees" 10 V, so each of the lamps will still draw 1 amp. The blow dryer will simply be drawing more. That is my understanding. So why do the lights dim?

Of course I realise my model could be wrong. I'm assuming a blow dryer has a fan and some heating elements. If the heating elements are anything like those tungsten filaments in toasters or whatever, then they heat up and glow because they have *high* resistance, not low right? So maybe my model is bogus. I still don't see what difference it makes to the lamps though.

meh, this post is already much longer than I thought it would be, so I will wait for your insights on my first question, and come back to ask the second later.

2. Jun 11, 2005

Averagesupernova

Have you considered the possibility that the source has an impedance? Although it is low, it is not perfect.

3. Jun 11, 2005

cepheid

Staff Emeritus
No, but I ought to have! If that is the simple reason, then let me talk it through. In class we primitively modelled "non-ideal" (that is to say, real) sources as an ideal voltage source in series with a small resistor. So, the more current that is drawn from the source, the lower the voltage it is able to provide. Say it is now 8 V, or whatever. You're still drawing more current than before, overall, but less through each branch than there would be if the blow dryer were not connected in parallel. So am I understanding you correctly? That's all there is to it? (neglecting other aspects of "impedance", like inductance or whatever, since my brain can't handle complicated cases :tongue: )

My second question is, what is meant by the 1600 W, exactly? The average power dissipated? Is it safe to say that ideally, the blow dryer draws 1600/120 = 40/3 = 13.3 A (RMS value)? Because the source is an AC source. 120 V is the RMS voltage, so the average power divided by this voltage gives RMS current, right? My understanding is that the RMS current is such the equivalent DC current in that load would disspate the same power as the average power dissipated in the AC case.

Also, does that sinusoidally varying power ever dip below zero with these typical household appliances? If so, does that mean that at a certain part of the cycle, the fortuitous combination of voltage and current in the load actually generates power? But on the other hand, maybe that's impossible. If it's only a resistive load, then the voltage and current are always in phase. This always confused the hell out of me. I have no problem manipulating phasors to mathematically sort things out, but interpreting what is going on, and what all that active/reactive power jargon really means is hopeless. I'm only wondering whether any of it applies to this situation...

4. Jun 12, 2005

Averagesupernova

1600 watts means that the appliance consumes 1600 watts. Not sure what you mean by the last paragraph.

5. Jun 12, 2005

cepheid

Staff Emeritus
If it is an AC voltage, the power consumption cannot be constant. I was simply asking whether it represents the average power consumed. Nevermind the last paragraph. I think I was off the wall on that one.

6. Jun 12, 2005

Gokul43201

Staff Emeritus
The resistance of the blow dryer is 125^2/1600 ~ nearly 10 ohms

Generate power across a resistor ? The heat produced (power dissipated) in a resistor comes from inelastic collisions...that does not change with AC. Besides, the resistive drop is in phase with the current, so there's no need to worry.

Last edited: Jun 12, 2005
7. Jun 12, 2005

Averagesupernova

I guess it's pretty obvious that the power consumption is not constant because it is AC. But power is just about always measured and stated as an average. RMS voltage muliplied by RMS current produce AVERAGE power.

8. Jun 13, 2005

cepheid

Staff Emeritus
Yeah, I realised that later. Thanks for stating it outright.

9. Jun 13, 2005

cepheid

Staff Emeritus
Yes! Again, thanks for confirming/stating it outright. It's what I was trying to get at in my second post, but it was kinda convoluted, I realise that. It's cleared up now. I get it.

10. Jun 13, 2005

SGT

Although we say that the voltage is 120Vrms, it is in reality 127Vrms between a phase and the ground and 220Vrms between phases. $$127 = \frac{220}{\sqrt{3}}$$
1600W is the total consumed power: resistive power from the heating and reactive power from the motor.

11. Jun 13, 2005

Gokul43201

Staff Emeritus
Forgot about the motor, didn't I ?

12. Jun 15, 2005

SGT

A correction of my post. 1600W is active power, both from the heating resistor and the motor.
An ideal motor with no charge has only reactive power. A real motor dissipates energy in the windings resistances even with no charge.
When the motor has a charge, like friction in the bearings and air resistance from the blower, the voltage and the current are no more at 90 degrees and there is active power consumed, otherwise we would have work with no expenditure of energy = perpetual motion.

13. Jun 15, 2005

Staff: Mentor

I didn't read the whole thread, so sorry if somebody said this already. The lights dim when you start up the motor because it has a very low input impedance when it is not spinning, or when it is heavily loaded. The lights return to almost normal brightness when the motor is spinning. If you stick a motor solid and apply power, you will generally burn out the motor with all the extra current that it draws when static.

14. Jun 25, 2005

Rogue Physicist

I can't help commenting here either. When anyone first learns about electricity and ohm's law, they think literally, just as they are taught. The wires are considered to have no resistance, the resistors are ideal etc.

And one of the first things a young experimenter discovers and has to wrestle with is the fact that no matter what load he puts on a wall-socket, the voltage stays at 120 volts a.c. or whatever, completely unhinging his understanding of Ohm's law with batteries and D.C. currents and voltages.

Then just as he gets over that ridiculous confusion, the lights dim when he plugs in a motor or heater, and he can't find the explanation.

But it turns out the mystery is not so difficult: Between the fuse-box or breaker panel and the wall is anywhere from 50 to 200 feet of copper wire, usually 12 or 14 gauge (as electrician's classify it), meant to carry a total of about 15 to 20 amperes safely, but not without a simple resistance based upon length of wire, and quality of connections in the junction boxes.

When the motor pulls the extra current down, the resistance between the fuse-box and the outlets, and the point of tie-in for the lights, as well as the size of service in the building and the wires running from the high-voltage transformer outside to the building-panel all must be considered, and explain why some lights dim, and others don't when a standard fused (and gauged) line is loaded down. Anywhere there is a resistance and a change of current, there is a voltage-drop.

Mystery solved. Ongoing dimness has nothing to do with A.C., or motors versus light-bulbs, but rather voltage-drops across supply-lines when current changes.

This can cause initially confusing, but not really mysterious effects of all kinds, like differences between grounds of up to 20 volts, which can give musicians a nasty shock when touching their grounded guitars while trying to swallow a microphone plugged into a unit grounded on another line.

Last edited: Jun 25, 2005