Resistors

Reply

  #1  
Old 01-27-03, 03:18 AM
josh1
Visiting Guest
Posts: n/a
Resistors

Hi

Im trying to work out some calculations and want to make sure Im on the right track.

Using Ohms law Voltage =resistance*amps

where AC is the current at 120 volts and I have an AC device that draws 15 watts.

So 120 volts= resistance* .125A ( 15w/120 volts)

So 120/.125= resistance

960= resistance

so the resistance of the device is 960 Ohms.

If I want to make the device draw 7.5 watts, I would substitute that for A draw.

120=resistance*.0625A (7.5w/120volts)

120/.0625= resistance

1920=resistance

so 1920 ohms resistance total in circuit

If resistance total= resistance of device+added resitor

1920=960 ohms + added resistor

1920-960= added resistor

960 = added resistor

How much variance should I account for in resistors? I was thinking 10% so my resistor range would be from 874-1056 Ohms.

Does this sound right? Are all resistors AC capable? Do I need a power factor correction in here for AC, because it is a very small draw, I was thinking it wont matter?

Thanks for the help-Josh
 
Sponsored Links
  #2  
Old 01-27-03, 05:37 AM
Member
Join Date: Nov 2001
Location: Taylors, SC
Posts: 9,483
Received 0 Votes on 0 Posts
If I understand you.....you have a circuit with a load of 15 watts and you want to make it draw 7.5 watts.

If that is the case, then you must reduce the load, not increase it.

Resistors dissipate power or use power. A load, such as a motor or light bulb, dissipates power or uses power.

If you have two light bulbs wired in parallel. Each one draws 7.5 watts of power. Total load on the circuit is 15 watts. In order to reduce the load, remove one of the light bulbs. Now the load is 7.5 watts.

Reducing the voltage will not matter, because the load will draw greater current to make up for the reduced voltage, thereby having the same load.

960 ohms load will dissipate 7.5 watts in your circuit.

Resistors are generally 10 - 20 %. Precision wound resistors are more on the order of 1 -10 %. For that matter, voltage supplied to your house has some variation to it as well.
 
  #3  
Old 01-27-03, 06:30 AM
josh1
Visiting Guest
Posts: n/a
Chris thanks for the help,

I believe what I need is a transistor, not a resistor as I thought.

Lets say I have a 120V circuit with a 15w light bulb. I want a three position switch, normally off, ON @15 W, ON @ 7.5 W. The light bulb has the same load draw all the time. Will a transistor vary Current (amps) to allow me to run the light bulb at 7.5W?

Thank you again-Josh
 
  #4  
Old 01-27-03, 07:22 AM
texsparky
Visiting Guest
Posts: n/a
Thumbs down Whats the motive?

Wouldn't it be easier to buy a dimmer that is U.L. listed instead of something homemade that may or may not be a fire hazard ?
 
  #5  
Old 01-27-03, 07:37 AM
Member
Join Date: Sep 2000
Location: United States
Posts: 18,497
Received 0 Votes on 0 Posts
First of all, a transistor has nothing to do with this. Other than the fact that both words end with "sistor", the two devices are not very similar.

Second, the whole plan may be a bad idea, depending on the setting. If this is a permanent installation, then the dimmer is a much better idea. But if this a science project, you can probably do this if you carefully monitor it. You'll only be dissipating 3.75 watts in the resistor, so as long as the resistor is rated to dissipate that many watts, you'd be okay.

Finally, your original analysis is essentially correct, assuming your load is purely resistive. A additional 960 ohm resistor in series with the light bulb will indeed cut the current in half, and reduce the power draw to 7.5 watts. Of course, the light bulb itself will dissipate only 3.75 watts, and the resistor will dissipate the other 3.75 watts.
 
  #6  
Old 01-27-03, 08:12 AM
josh1
Visiting Guest
Posts: n/a
So if I want the light bulb to draw 7.5 watts, I need what size resistor? How do I caluculate that? so that the total current draw is 15 watts, with 7.5 being used by the resistor, and 7.5 by the light. it seems to me like i need only 1 ohm resistor? that cant be right? Does the resistor always use 1/2 the current that it is reducing, or is it proportional to its resistance?

This is a "project" not a hardwire installation, not "really" using light bulbs, but it helps me understand.


Thanks for the help -Josh
 
  #7  
Old 01-27-03, 08:14 AM
RickJ6956
Visiting Guest
Posts: n/a
You need a fairly meaty resistor. Those carbon-based things you find at Radio Shack are usually 1/4 or 1/2 watt at 10% tolerance. If you install one of them in your circuit it will act like the burner on an electric stove.

You'll need at least a 25 watt resistor to handle the turn-on surge. A ceramic resistor is probably your best choice for dissipating power. Remember, though, that any resistor will dissipate power in the form of heat. It needs to breathe to cool itself.

Tolerances range from 20% to 1%. The cost goes up as the tolerance goes down. You'll easily get by with a 10-percenter.

Technically, a resistance measurement on an AC circuit is not going to give you real-world load consumption. "Impedance" is what you need to measure. But since you're only using the resistor to dim a light bulb, it's close enough.
 
  #8  
Old 01-27-03, 08:16 AM
RickJ6956
Visiting Guest
Posts: n/a
Woah! Just saw your latest post. If not light bulbs, what are you hooking up? You can burn out certain motors ...
 
  #9  
Old 01-27-03, 09:05 AM
Member
Join Date: Sep 2000
Location: United States
Posts: 18,497
Received 0 Votes on 0 Posts
You cannot use a light bulb for the theoretical exercise if you're not going to use a light bulb for the project. The answer will not be correct for your actual load is not purely resistive.

There is no way to put a resistor in series with a 15-watt light bulb such that each uses 7.5 watts.

When the 15-watt light bulb is alone on 120 volts, it draws 0.125 amps. To get it to act as a 7.5-watt bulb, we need to decrease the current to 0.125*sqrt(2)/2, or 0.0884 amps. To get 0.0884 amps with 120 volts, we need a total resistance of 120/0.0884, or 1358 ohms. We already have 960 ohms from the light bulb, so we need to add a series resistor of 398 ohms. The resistor will then cause a voltage drop of 0.0884*398, or 35 volts. That means that the resistor will dissipate 35*0.0884, or 3.1 watts. The light bulb will cause a voltage drop of 85 volts, and dissipate 7.5 watts, which was the objective.

But as I said, this will not work if your real load is not purely resistive.
 
  #10  
Old 01-27-03, 09:26 AM
Member
Join Date: Dec 2000
Posts: 510
Received 0 Votes on 0 Posts
Assume that you have a purely resistive load, call it RL. When connected directly across the 120V input voltage it dissipates 15W. You want to add a resistance is series with RL so that the dissipation in RL drops to 7.5W.

Power in a resistive load equals V^2/R. If at 120V RL dissipates 15W, then RL equals 960 ohms (your answer). In order to reduce the dissipation in RL to 7.5W, the voltage across RL must be reduced to ~85V.

The voltage can be reduced by adding a series resistor RS. For RS and RL in series, the voltage across RL is V*RL/(RS+RL). Solving this equation to find RS gives:

RS = RL*(Vin - VL)/VL

where Vin is the input voltage (120V) and VL is the desired voltage across RL (85V in this case). Plugging in the values gives a result of ~395 ohms for RS.

Note that RS will dissipate significant power. The voltage across RS is 120V - 85V = 35V. The power dissipated in RS is then 3.1W.

Check: Total of RL plus RS is 1355 ohms. 120^2/1355 give 10.6W power dissipated in the two resistors. 7.5W + 3.1W = 10.6W so the answer checks.
 
  #11  
Old 01-27-03, 09:29 AM
josh1
Visiting Guest
Posts: n/a
John you are a genuis! I was trying to work it out keeping the voltage drawn at 15, which clearly is why I failed physics. How did you get the sq rt(2)/2 part though? I owe you many drinks if you are ever in PA.

Thanks again-Josh
 
  #12  
Old 01-27-03, 09:36 AM
brickeyee
Visiting Guest
Posts: n/a
filament light bulbs (tunsten, halogen, xenon) do not have enough inductance to create any significant reactance at 60 hertz. They are a purely resistive load. The problem that occurs is that the resistance of a cold filament is significantly less than of a hot filament. This has blown many a solid state device trying to switch a filament on. If you measure the resistance of a light bulb and try to compute power from this, it is the change in resistance with temperature that gives the incorrect answer, not any reactive load.
 
  #13  
Old 01-27-03, 09:38 AM
Member
Join Date: Sep 2000
Location: United States
Posts: 18,497
Received 0 Votes on 0 Posts
Original: P1 = I1^2 * R
Desired: P2 = I2^2 * R, where P2 = P1/2

Thus, I2^2 = 1/2 * I1^2
Which yields I2 = SQRT(1/2) * I1

Note that SQRT(1/2) = SQRT(2)/2
 
  #14  
Old 01-27-03, 09:55 AM
Member
Join Date: Nov 2001
Location: Taylors, SC
Posts: 9,483
Received 0 Votes on 0 Posts
parallel and series can be confusing. at least we got him on the right track.
 
  #15  
Old 01-27-03, 10:03 AM
josh1
Visiting Guest
Posts: n/a
I just calculated it out for 30 watt device with a 15 watt output.

The resistor would need to be 198 Ohms. The resistor would dissipate 6.2 watts and the device would dissipate 15. The total watts dissipated 120^2/678=21.2W (where device R=480 and resistor value=198 so Rtotal=678ohms)

I checked it using mikes method, very helpful. I was missing too many formulas to understand this earlier.

Thanks for all the help-Josh
 
  #16  
Old 01-28-03, 12:44 PM
Member
Join Date: Feb 2002
Location: port chester n y
Posts: 2,117
Received 0 Votes on 0 Posts
Connect is "series" across 120 (+,-) volts--------two 15 watt lamps and a muti-meter with a 0-10 amps AC range. Muliply the amp reading X the voltage reading for the total power in watts and divide the wattage by 2 for the power consumed by each lamp.Divide the voltage reading by the amps reading to calculate the total resistance of the circuit.
 
  #17  
Old 01-29-03, 12:39 PM
jjl3rd
Visiting Guest
Posts: n/a
Interesting problem.
Another way to obtain the same result is to eliminate 1/2 of the voltage. With AC this is easily accomplished by removing 1/2 of the waveform. Place a diode in series with the line. That way only the positive ( or negative depending on the diode orientation) 1/2 cycle is present at the load. The advantage is that a relatively small and cheap diode can be used when compared to the high wattage resistor required.
john
 
  #18  
Old 01-30-03, 01:09 AM
josh1
Visiting Guest
Posts: n/a
a diode will be black with a silver stripe and it has to orient a certain direction toward the draw correct? Thanks-Josh
 
  #19  
Old 01-30-03, 06:46 AM
jjl3rd
Visiting Guest
Posts: n/a
Yes that is a ok description of a diode. An easy to find diode that will work for the application mentioned would be a 1n4004 which is a diode capable of 1 amp. and good to I think 400 volts. also good are the 1n4005, 1n4006, and 1n4007 that are all 1 amp devices just able to withstand more voltage. I think radio shack even has them by that number. If you try to substitute a different one remember that the voltages we usually work with are RMS voltages and the peaks are higher (1.414 times higher plus any line noise) so use at least a 400 volt rated diode.
john
 
  #20  
Old 01-31-03, 08:48 AM
RickJ6956
Visiting Guest
Posts: n/a
In this application it won't matter which way you orient the diode. All you're doing is chopping half the waveform. The lightbulb doesn't care which half it sees.
 
  #21  
Old 02-01-03, 04:55 AM
jjl3rd
Visiting Guest
Posts: n/a
Thanks Rick, I answered everything but his question.
 
Reply

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Thread Tools
Search this Thread
 
Ask a Question
Question Title:
Description:
Your question will be posted in: