Resistor power rating increase in parallel?

Let's say I need to have a resistor in a circuit rated for 10 Watts. If I had two resistors rated for 5 W, could I wire them in parallel to make the equivalent power rating 10 W? I know my resistance will be different but I'm not sure how the power rating would differ.

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The power would share. If you wanted, say, 50Ohms 2 Watt, and you had two 100 Ohm, 1 Watt, in parallel you'd get 50Ohm 2 W.

BUT if you had two 25 OHM 1W resistors in SERIES, you would ALSO have a 50 Ohm 2W resistor.

No, if you place them in series the wattage will not increase, only the resistance. The electrons passing though two equal resistors in parallel will have two paths to travel at the same time and 50% will pass though each resistor (effectively doubling the power rating of the equivalent resistor). Electrons passing through two resistors in series will only have one path to travel at each given time (through the first resistor, then the next resistor).

Sorry, you're wrong. Think harder next time.

Tell me this: Let I=0.2A Let R= 25 Ohms, and the limiting disspation 1W.

Put two in series.

I^2 x 2R = ???

I halved the resistance of the series case. If P=I2xR/2 +I^2R/2 Then P= I^2 (R/2+R/2)= I^2xR.

It is easy enough to just calculate the voltage and current through each individual resistor in your network, for to confirm that each individual resistor does not dissipate more power than its rating allows.

For example, if your original resistor had a value of 40 ohms, and was rated for 10 watts, that means the maximum voltage you could put across it would be 20 volts, since Pmax = Vmax^2/R = 20*20/40 = 10. Also note, the maximum current through this 40 ohm resistor is 0.5 amperes, since Pmax = I^2*R = 0.5*0.5*40

If you replace your original 40 ohm resistor with two 80 ohm resistors wired in parallel, the voltage across each resistor is the same, because they're wired in parallel. Like before, the maximum voltage is expected to be 20 volts. So the maximum power dissipated by each 80 ohm resistor is expected to be,

Pmax = Vmax^2/R = 20*20/80 = 5 watts. Also note, the maximum current through each resistor is expected to be (20 volts)/(80 ohms) = 0.25 amperes.

Notice how this arrangement of two identical resistors in parallel has divided the current neatly in half. Since both resistors share the same voltage, because they're in parallel, this arrangement also neatly divides the power dissipation in half, since power is voltage times current.

By the way, if you choose an arrangement of resistors that are not exactly the same size, the power dissipation will NOT be divided exactly evenly among the individual resistors. If the resistors in your network are approximately the same size, the power will be divided in a way so that each individual resistor dissipates approximately the same amount of power, if that makes sense.

seandogue1 year ago

Yes. In brief, power rating doubles for a parallel connection of two resistors of equal value and power rating. The individual resistors will need to have twice the resistance of the required value [ Reff = R1*R2/(R1+R2) ].

One other thing. Resistors have tolerances, ie, 1%, 5%, 10%, etc.

For two random 10% resistors, this means that in the worst case, one resistor will be 90% of the desired resistance, the other 110%.

If you operate at the edge (ie, suppose you want to deliver ~20W of power, then the lower valued resistor will draw 55% of the current and may burn out if you drive the pair to its mathematical limits

If you have a large supply of resistors from which to pick, you can mitigate this caveat (to a point) by hand selecting the resistors, measuring each until you've found two that match in their resistance values. If not, then you must perform a worst-case analysis to determine the maximum power you can safely deliver with the pair.

Do not select my answer or I will be forced to delete it.

Yes you can run 10 watts on two 5 watt resistors as long as the resistance of the two resistors are the same. the R total of the circuit will be 1/2 and strangely enough the resistance is more accurate as it halfs the tolorence.