Resistances of common appliances


by jslatane
Tags: examples, resistance
jslatane
jslatane is offline
#1
Sep25-10, 05:41 PM
P: 11
Hello, this is my first post on Physics Forums, so if I am making any mistakes, please let me know!
My reason for this post is that I am giving a speech to a small class in a few days about resistors. What I wanted to do was to calculate the resistance of some real life objects that people are familiar with, just to sort of show them that resistance could be used to describe things they interact with all the time. So I figured I would start with a bare wire, the resistance of which is about as close to zero as physically possible, then the resistance of an LED, then a normal lightbulb, small appliance, large appliance, and gradually work my way up to say a house, a town, a large city, a country, etc, giving the resistances of each.
SO I started doing some calculations using numbers I found on various websites. Here they are:

E = P*T
P = I*V
R = P/(I^2)
I = P/V
R = P/(P^2/V^2)
______________________________
For bare wire:
R = ~0 ohms
For LED:
V = 3.2 volts
I = 20 mA = 0.02 amps = 2/100 amps
R = V/I = 3.2/0.02 ohms = 3.2*100/2 ohms = 320/2 ohms = 160 ohms
For standard incandescent lightbulb:
P = 100 watts
V = 120 volts
I = P/V = 100/120 amps = 1/1.2 amps = 0.833 amps
R = V/I = 120/0.833 ohms = 120*1.2 ohms = 144 ohms
For microwave:
P = 1030 watts
V = 120 volts
I = P/V = 1030/120 amps = 1.03/0.12 amps = 8.5833 amps
R = V/I = 120*0.12/1.03 ohms = 13.98 ohms

And that is as far as I got before I started wondering what was going on. For some reason I had the idea that as a device got larger, the resistance would increase. But this is not what I am seeing. Is that an unrealistic expectation? Are my calculations flawed? Are there better ways to show resistances of everyday objects? Please note that I am not asking for specific numbers or answers, thus I didn't post it in the homework section, but I can move it if need be. I would appreciate it if someone could just explain to me if my examples are valid and my methodology is realistic or flawed.
Thanks!
Phys.Org News Partner Engineering news on Phys.org
Lifting the brakes on fuel efficiency
PsiKick's batteryless sensors poised for coming 'Internet of things'
Researcher launches successful tech start-up to help the blind
vk6kro
vk6kro is offline
#2
Sep25-10, 07:01 PM
Sci Advisor
P: 4,003
No, as each device draws more power, at a constant voltage, the current must be increasing and the resistance must be getting smaller.

Power = voltage * current

Current = voltage / resistance

so Power = voltage * voltage / resistance

So, if the power goes up and the voltage stays the same, the resistance must decrease.
jslatane
jslatane is offline
#3
Sep25-10, 07:50 PM
P: 11
Ok, that makes sense, I'm not sure why I wasn't looking at it like that before. So what about for larger systems? Say I wanted to describe a household, or really all the electronic devices within it. A quick google search gives me a ballpark figure of 10,000 kWh for an average US home. If the power lines transmit electricity at the standard 120 volts to the house, then
energy = power*time
10,000 kW = 10,000K kWh / 1 hour
power = voltage^2/resistance
10^6 W = 14,400 V/resistance
resistance = 14,400/10^6 ohms = 0.0144 ohms
So let me see if I understand this correctly: resistance is the measure of how much the material opposes a current flow. Ok..
So basically what I'm getting from this is that trying to compare just resistance really doesn't mean much unless you describe the circuit it is in.

vk6kro
vk6kro is offline
#4
Sep25-10, 09:35 PM
Sci Advisor
P: 4,003

Resistances of common appliances


Yes, that seems OK except for this typo:

resistance = 14,400/10^6 ohms = 0.0144 ohms

.....should be 10^7 watts or 10 000 000 watts for 10000 KW
so you would get 0.00144 ohms

1 watt = 14400 /1 = 14400 ohms
100 watts = 14400 /100 = 144 ohms
1000 watts = 14400 / 1000 = 14.4 ohms
10000 watts = 14400 / 10000 = 1.44 ohms
50000 watts = 14400 / 50000 = 0.288 ohms

However, this is going to deteriorate into some very small numbers as you go up in power.
Maybe it would be better to talk about current, or you could lose your audience as they get more confused.

eg 50000 watts at 120 volts is 50000 / 120 = 416. 67 amps
That is a lot more impressive than saying 0.288 ohms.
aks786
aks786 is offline
#5
Sep26-10, 08:35 AM
P: 28
Also, such higher loads (10,000kW) are never operate at low voltage (120V) because of practical issues (losses in connecting wires). they are fed on higher voltages and/or preferably at 3-phase system.
jslatane
jslatane is offline
#6
Sep26-10, 09:17 AM
P: 11
Thanks you guys for the corrections and suggestions. I think I will be able to better explain now that the resistance of an object depends on the situation it is in, and generally as power consumption of the system goes up, the system's resistance will decrease.
I will probably not address this in my speech, but a 3 phase system is just for AC right? And the main reason for it is so you get a fairly constant current versus a peak and a trough basically?
Averagesupernova
Averagesupernova is offline
#7
Sep26-10, 10:29 AM
P: 2,451
I think it is easy to think of resistance, or technically impedance, as the ratio between voltage and current. Obviously ohms law states it this way but I don't think most laypersons think of it this way. They think of it as some property of a conductor or semiconductor that opposes the flow of electrons. I know many people will disagree with me but a water analogy does help people understand. We need to keep in mind that no analogy is perfect though.
jslatane
jslatane is offline
#8
Sep26-10, 10:45 AM
P: 11
Yeah, that is basically the conclusion I reached from this discussion. Initially I kind of fell into the thinking that you describe, as though resistance is opposition to flow of electrons, but now I realize it does not affect the current, as in the current coming out is not any less than what goes in, but it does affect the voltage, correct? I was also planning on using the water analogy in my speech just for voltage, current and resistance so I am glad you mentioned this. Thanks
vk6kro
vk6kro is offline
#9
Sep26-10, 08:13 PM
Sci Advisor
P: 4,003
Quote Quote by jslatane View Post
Yeah, that is basically the conclusion I reached from this discussion. Initially I kind of fell into the thinking that you describe, as though resistance is opposition to flow of electrons, but now I realize it does not affect the current, as in the current coming out is not any less than what goes in, but it does affect the voltage, correct? I was also planning on using the water analogy in my speech just for voltage, current and resistance so I am glad you mentioned this. Thanks
Resistance IS the opposition to flow of electrons and the voltage does not change, so I wonder if you have it right?

Imagine what happens in your house. The voltage coming in is very steady and stays the same no matter what appliances you turn on.
You can tell this by watching the lights when you turn on a clothes dryer or something that uses a lot of power. They stay at the same brightness, meaning they are getting the same voltage.

As you turn on each extra thing, it takes electrons from the fairly constant voltage coming in from the street. This is electric current. It goes from one wire of the supply, through the appliance or light and back to another wire of the supply.

As you turn on more and more of these lights or appliances, the total current keeps increasing. For the purposes of this discussion, the total resistance keeps going down.
jslatane
jslatane is offline
#10
Sep26-10, 08:30 PM
P: 11
I didn't explain myself very well, I meant that voltage drops across a resistor. This is what happens, correct?
vk6kro
vk6kro is offline
#11
Sep26-10, 08:53 PM
Sci Advisor
P: 4,003
It might just be how you are saying it.

The resistor doesn't cause the voltage to drop. It is placed across two conductors which already have different voltages on them, so there is a voltage difference between them.

Because of this voltage difference and the resistance of the resistor, there will be a current flowing.
jslatane
jslatane is offline
#12
Sep26-10, 08:58 PM
P: 11
Well how can you have a voltage without resistance? You say that the resistor is placed across two conductors which already have different voltages on them, which implies that the voltage would be there even if the resistor was not. But V = IR and if R is 0, then there is no voltage, is that not correct? So there must not be a voltage difference until you place the resistor into the circuit, correct?
vk6kro
vk6kro is offline
#13
Sep26-10, 09:11 PM
Sci Advisor
P: 4,003
R is not zero is it? It is infinite.

There certainly is a voltage even without current. The power company supplies a voltage to your house and makes sure that when you draw current from it (within limits) this voltage stays fairly constant.

As you connect more lights across this supply, by switching them on one at a time, then more current is taken from the supply.
jslatane
jslatane is offline
#14
Sep26-10, 09:13 PM
P: 11
Resistance is infinite? Wouldn't that make voltage infinite as well unless current is zero?
vk6kro
vk6kro is offline
#15
Sep26-10, 09:29 PM
Sci Advisor
P: 4,003
But the current IS zero. You have a light socket with no lamp in it. What current is flowing?

No, the voltage is constant. It is 120 volts AC (in your case) and this does not vary.

When you put a resistance across it, there will be a current flowing in accordance with Ohm's Law which is I = E / R.
jslatane
jslatane is offline
#16
Sep26-10, 09:32 PM
P: 11
Ok, that makes sense. Does E stand for voltage in that equation? Thanks for being patient with this.
Zryn
Zryn is offline
#17
Sep26-10, 10:00 PM
PF Gold
P: 322
E stands for Electromotive Force, another name for Voltage.
jslatane
jslatane is offline
#18
Sep26-10, 10:20 PM
P: 11
Oh ok, I had forgotten about this term, thanks.


Register to reply

Related Discussions
I Hate modern appliances! General Discussion 17
Common collector vs Common drain amplifier Electrical Engineering 7
common emitter vs. common collector power amplification Electrical Engineering 3
CCMRR Common to Common Mode Rejection Ratio Electrical Engineering 3
AC and Appliances Electrical Engineering 12