
Author  Message 

andyfrogs@aol.com
Posts : 3 Join date : 20101017
 Subject: Added wattages rule? Sun Oct 17, 2010 6:56 pm  
 Hi, How does it work? Do wattages add in series or in parellel? Say I want to equal 4 watts. Will two 2 watt resistors in series do it? Also, is there a visual chart on the wattage sizes of dog bone restistors? I have one that is about 2 inches long. I think that is a 2 watt? But, I am not sure. Please refresh my memory! I am sorry to hear about the latest problems this great site is having. I hope we never loose this forum!!!
Thanks, Andy andyfrogs@aol.com 

 
jack shirley
Posts : 67 Join date : 20101012 Location : SE USA
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 7:21 pm  
 There's a good page on Chuck Schwartz's Philco site that shows photos of new vs old resistors and their wattage. Sri, just changed machines and I can't give you a quicky link.
You can find an infinite number of resistor combinations to equal what you need but at the end of the day its probably less than a buck just to order a proper component from mouser or whoever. 

 
radiotechnician
Posts : 35 Join date : 20101012
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 7:31 pm  
 If you had two 10 watt 20 ohm resistors you could connect them in series to get a 40 ohm 20 watt resistor.
If you had two 2 watt 22 ohm resistors you could connect them in parallel to get an 11 ohm 4 watt resistor.


 
Carl_Travis
Posts : 8 Join date : 20101016
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 7:36 pm  
  jack shirley wrote:
 There's a good page on Chuck Schwartz's Philco site that shows photos of new vs old resistors and their wattage. Sri, just changed machines and I can't give you a quicky link.
Here's the link to the pics: http://www.philcorepairbench.com/resistcompare.htm 

 
Dale Saukerson
Posts : 6 Join date : 20101012
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 7:55 pm  
 The wattage ratings don't add up. You have to calculate the wattage seen by each resistor, either in series or parallel.
Let's say the original resistor is 20 Ohms at 10 Watts.
This can be replaced with two series resistors, each 10 Ohms and each rated at 5 Watts. You can NOT use a 10 Ohms @ 2 Watts in series with 10 Ohms at 8 Watts.
You can replace it with two 40 Ohm @ 5 Watt resistors in parallel. You can NOT replace it with a 40 ohm @ 2 Watt in parallel with 40 Ohm @ 8 Watt
There is no "rule" to figure this out. The math is easy when the separate resistors have identical Ohmic and Wattage values. If they do not have identical values, then Ohms law is your friend. 

 
OtakuN3rd
Posts : 18 Join date : 20101013 Age : 28 Location : Dowagiac, MI
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 10:13 pm  
 I had learned that in a group of resistors you teat the wattage they're capable of as whatever the lowest watt rating is.
I could be wrong, though... 

 
oldradioparts
Posts : 23 Join date : 20101013
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 10:20 pm  
  OtakuN3rd wrote:
 I had learned that in a group of resistors you teat the wattage they're capable of as whatever the lowest watt rating is.
I could be wrong, though...
You are indeed wrong. The wattage calculation has to do with the current and voltage drop across each resistor. If both resistors in a series OR parallel configuration are equal resistance, its easy to add the wattages for a total wattage capability. But, when the resistances are unequal, unequal currents run through them, so the wattage needs are unequal. Mark Oppat www.oldradioparts.net 

 
OtakuN3rd
Posts : 18 Join date : 20101013 Age : 28 Location : Dowagiac, MI
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 10:42 pm  
 That makes sense now that you mention it. Just gotta remember the P=I*E formula... mmmm.... pie 

 
Radiosmoker
Posts : 20 Join date : 20101014 Age : 73
 Subject: Re: Added wattages rule? Sun Oct 17, 2010 11:29 pm  
 You cannot have a resistor that is underrated. If the entire circuit expends 10 watts and you have one resistor that is designed for 1 watt and a second series resistor that is 10 watts, You cannot force 10 watts of power to be dissipated by the 1 watt resistor.
Because the voltage drop on each resistor is the same. In a series circuit the resistor that is the highest resistance will have the most voltage drop. Since the resistances are equal they have to share the load equally.
Think of it as a two short pieces of wire hooked in series but one is 32 gage and the other as 0000 gage. For a short length of wire we will argue that the resistance is so small that its not worth the trouble of arguing over. We send 100 amps of current though the series wires. What happens? Yes for a fraction of a second 100 amps will flow, but then what happens in the next second? The smaller wire will glow red hot and evaporate. Yet the wire did handle the power for a fraction of a second. The smaller wire could not dissipate the total current forever, so destruction of the wire took place.
(or think of two lightbulbs in series = what happens?) Both light bulbs are 25 watts, question will the lights both have the same brilliance in a series circuit as the brilliance of one alone?
Same takes place with resistors = current is limited through a series circuit by the highest resistance and the same with wattage.
Resistors in parallel is another story. 

 
oldradioparts
Posts : 23 Join date : 20101013
 Subject: Re: Added wattages rule? Mon Oct 18, 2010 12:07 am  
 for equal resistances, wattage dissipation will be equal. BUT, Say you need 10watts dissipation, and the total resistance needed is 10K. You can have a 1 watt 1 K resistor and a 9 watt 9K resistor in series. The wattages add, so do the resistances. But, you cant have a 1 watt 5K resistor and a 9watt 5 K resistor. Remember, equal resistances must have at least half the wattage rating of the total needed.
In parallel, its more complex because the resistances dont add simply. If you are using equal resistances, its easy again, as the wattage dissipation requirement will be equal (again, you can always use a higher wattage part for either or both resistors if you have one). So if you need a 50ohm 20 watt resistor, then two 100ohm 10 watt types in parallel do the job. for just two resistors in parallel the law is R1 + R2 / R1 X R2 You then have to figure out the wattage dissipation on each resistor using P (power) = E squared X R
Mark Oppat www.oldradioparts.net 

 
andyfrogs@aol.com
Posts : 3 Join date : 20101017
 Subject: Re: Added wattages rule? Mon Oct 18, 2010 6:54 am  
 Hi, Ok, the easiest thing for me to keep in mind is. As long as the resistances are the same the wattage will disipate the heat equally. What I am trying to do here is replace the 2 watt dog bone that should be 13K ohms,but,is more like 23K ohms with 2 6500 ohm 2 watt carbon composition resistors in series. I guess 2 6500 1 Watts in series would work to? Am I right?
Thanks, Andy 

 
Dale Saukerson
Posts : 6 Join date : 20101012
 Subject: Re: Added wattages rule? Mon Oct 18, 2010 12:13 pm  
 Andy, "yes" to your last question.
2 @ 6500 Ohms @ 1 watt, in series, equals 13000 @ 2 watts 2 @ 6500 Ohms @ 2 watt, in series, equals 13000 @ 4 watts
In both cases, the wattage dissipation capability will be equally shared across each resistor. This is true only because the resistors are of identical Ohmic value.
If their Ohmic values were not equal, you would need Ohms law to calculate how much wattage each resistor would see in the circuit being used. 

 
andyfrogs@aol.com
Posts : 3 Join date : 20101017
 Subject: Re: Added wattages rule? Mon Oct 18, 2010 12:46 pm  
 Hi Dale, Thank you for a very clear explaination to my question. I am greatful to all that contributed. I am working on a 1942 Crosley 83CP console.
Thanks, Andy 

 
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