Hello,

If two 20 watt resistors are wired in parallel what will the wattage be?

If the same but in series?

Mitch

40 Watts in series or parallel.

Two identical resistors in parallel will have half the resistance of one, and twice the wattage rating. Two resistors in series will have twice the resistance of one, and twice the power rating.

Assuming that they are the same resistance value, when wired in parallel the power will be split equally between the two parts and the maximum total power capability will be 40 watts.

If the resistors are unequal resistances, the power handling capability of the two will be more than 20 ohms but less than 40 ohms.

For series connected resistors of any ohmic value (need not be the same) the max power handling capability if both resistors are rated for 20 watts is... 20 watts.

:Hello,
:
:If two 20 watt resistors are wired in parallel what will the wattage be?
:
:If the same but in series?
:
:Mitch
:
Mitch : 40 watts total regardless of how you wire them . 20 watts each developed by the currents that go through them and the voltage applied.The current and voltage depends on what you hook them to and how.
Plus as in most manufactured stuff there is a saftey factor. But 20 watts is the continous rating.It is not a problem to use an oversized resistor usually. Unless you do not have space or the cost is high . When you get into these larger sizes they are often wire wound. Those are large resistors ,but maybe you need larger ones ? At anyrate these will put out some heat if they are at rated wattage. Normally these radios are around 100 watts total .40 watts would be some heatif under a chassis.
bill

:Hello,
:
:If two 20 watt resistors are wired in parallel what will the wattage be?
:
:If the same but in series?
:
:Mitch
:
Mitch : If you are replacing a series wired 40 watt resistor with two parallel 20 watt resistors in a circuit that is a different quuestion.Sort of. You would have to make the parallel resistance correct or equal to the original.So it would be a matter of getting the resistances right (by formula).That would be easy to figure .(R1*R2)/(R1+R2)
The power in the two branches would still be capable of 40 watts .The actual power would be the sum of the power in the two branches.Hope that helps.
bill

::Hello,
::
::If two 20 watt resistors are wired in parallel what will the wattage be?
::
::If the same but in series?
::
::Mitch
::
:Mitch : If you are replacing a series wired 40 watt resistor with two parallel 20 watt resistors in a circuit that is a different quuestion.Sort of. You would have to make the parallel resistance correct or equal to the original.So it would be a matter of getting the resistances right (by formula).That would be easy to figure .(R1*R2)/(R1+R2)
:The power in the two branches would still be capable of 40 watts .The actual power would be the sum of the power in the two branches.Hope that helps.
:bill
:
Thank you everyone,

Perhaps I should explain.

I am replacing a line cord that has a resistor wire in it.

I have calculated the resistance and the voltage drop required to operate the filaments.

I need 310 ohms of resistance. I connected two 640 ohm 5 watt resistors in parallel to get 320 ohms. Does this give me 10 watts?

The radio worked but the resistor gets real hot.

I have some 100 ohm 10 watt resistors that I can use in series to get what I need.

If I use three of these 10 watt resistors will it be cooler?

Thank you,

Mitch

:::Hello,
:::
:::If two 20 watt resistors are wired in parallel what will the wattage be?
:::
:::If the same but in series?
:::
:::Mitch
:::
::Mitch : If you are replacing a series wired 40 watt resistor with two parallel 20 watt resistors in a circuit that is a different quuestion.Sort of. You would have to make the parallel resistance correct or equal to the original.So it would be a matter of getting the resistances right (by formula).That would be easy to figure .(R1*R2)/(R1+R2)
::The power in the two branches would still be capable of 40 watts .The actual power would be the sum of the power in the two branches.Hope that helps.
::bill
::
:Thank you everyone,
:
:Perhaps I should explain.
:
:I am replacing a line cord that has a resistor wire in it.
:
:I have calculated the resistance and the voltage drop required to operate the filaments.
:
:I need 310 ohms of resistance. I connected two 640 ohm 5 watt resistors in parallel to get 320 ohms. Does this give me 10 watts?
:
:The radio worked but the resistor gets real hot.
:
:I have some 100 ohm 10 watt resistors that I can use in series to get what I need.
:
:If I use three of these 10 watt resistors will it be cooler?
:
:Thank you,
:
:Mitch
:
Mitch : I am guessing that would spread the heat out and each one would be cooler . I have never dealt with the resistance cords.Someone else on this site must have worked around these so I'll stop here.Old
trf's did this voltage drop with a heater wire in a perforated can on top of the chassis . They put out some heat!
bill

Mitch,
It's not for nothing that these resistance cords were named "curtain burners". By using a pair of resistors instead of a line cord, you are concentrating all of the heat in a couple of parts instead of distributing it in the length of the power cord. So, expect it to get very hot.

Your radio probably was not designed to deal with this amount of heat inside the cabinet, so a bit of creative post-engineering might be needed to allow the modified unit to operate safely: such as some kind of heat sink/radiator.

:::Hello,
:::
:::If two 20 watt resistors are wired in parallel what will the wattage be?
:::
:::If the same but in series?
:::
:::Mitch
:::
::Mitch : If you are replacing a series wired 40 watt resistor with two parallel 20 watt resistors in a circuit that is a different quuestion.Sort of. You would have to make the parallel resistance correct or equal to the original.So it would be a matter of getting the resistances right (by formula).That would be easy to figure .(R1*R2)/(R1+R2)
::The power in the two branches would still be capable of 40 watts .The actual power would be the sum of the power in the two branches.Hope that helps.
::bill
::
:Thank you everyone,
:
:Perhaps I should explain.
:
:I am replacing a line cord that has a resistor wire in it.
:
:I have calculated the resistance and the voltage drop required to operate the filaments.
:
:I need 310 ohms of resistance. I connected two 640 ohm 5 watt resistors in parallel to get 320 ohms. Does this give me 10 watts?
:
:The radio worked but the resistor gets real hot.
:
:I have some 100 ohm 10 watt resistors that I can use in series to get what I need.
:
:If I use three of these 10 watt resistors will it be cooler?
:
:Thank you,
:
:Mitch

Instead of using resistors which produce heat, just use a capacitor. See this link http://www.nostalgiaair.org/References/Articles/TheFlash/Flash01.htm

Dave
:

You can also use a 1 amp ac diode in series with an 185 ohm 16.5 watt resistor. If you put any combination of resistors together you add the wattage for a total.

:You can also use a 1 amp ac diode in series with an 185 ohm 16.5 watt resistor. If you put any combination of resistors together you add the wattage for a total.
:
I can get a 1 amp diode at RS and I have the resistor combination.

Does the polarity of the diode matter?

Mitch

Polarity does not matter in this case. The filament string is just a loop in an AC line.

:Polarity does not matter in this case. The filament string is just a loop in an AC line.
:
:
:
Thank you Warren,

I will perform this task tomorrow. I do have a good space on top of the chassis to mount the resistors and diode for heat dissipation.

I do not understand this combination as to no polarity required for the filament voltage. Aren't I sending a half wave pulse to the filaments?

Mitch

PS
Dial cord repaired!

This site has all of your questions answered.
It is good practice to connect the AC line to the anode side of the AC diode.

http://www.vintage-radio.com/repair-restore-information/valve_dropper-calcs.html

To make the dial light work. Wire a 44 lamp in series with the voltage dropper. Add a 39 to 47 one watt resistor shunting the dial light.

:To make the dial light work. Wire a 44 lamp in series with the voltage dropper. Add a 39 to 47 one watt resistor shunting the dial light.
:
Thank you everyone,

I am using a 1 amp diode in series with a 200 ohm 20 watt resistor for the dropping resistor.

There is a factory installed 1200 ohm 20 watt resistor with a center tap to supply the dial lamp which is a 6 volt 150 ma bulb.

The resistor still gets hot but not like the 330 resistor I was trying to use.

There is space on top of the chassis where the 1200 ohm resistor is to mount the new resistor.

Thanks again gentlemen,

Mitch