If we have two resistors R1=10Ω and R2=20Ω , the both have same maximal allowed power Pmax=1kW. Find the maximal power of the series of these two resistors?

I thought i could do it this way:

since P=I^2*R so if i have Pmax i can find Imax for both cases:

Imax1=sqrt(P1max/R1)=+-10A
Imax2=sqrt(P2max/R2)=+-5*sqrt(2)

since the Imax2 is lower than Imax1 it means that maximum current through the series is 5*sqrt(2), since Re=R1+R2=30Ω
Pmaxe=(Imax2)^2*Re=50*30=1,5kW.

Is this good approach, or im missing something?

 

Hi,

Yes that seems to work. The larger value resistor sets the limit on the current and the current is the same in both resistors so the max through both must be the max for the larger value resistor. Knowing this, the total power dissipation is the sum of the power dissipation in each with the max current flowing in each one. That totals 1500 watts.

You might note however that as a practical application this might not work in real life because if they really are resistors then the 20 ohm resistor would get super hot because it is run at the maximum power, which is probably based on a 20 or 25 degree C ambient anyway. The better choice would be to run it at 1/2 the rated power, which would then allow it to stay cooler.
At the max current level the 20 ohm resistor would get super hot while the 10 ohm resistor would stay cooler because it would be run at only 1/2 it's rated power.
The exception is if they are heater elements, which you might want to run at the normal operating current.
To give a quick example, i have a resistor that is rated for 50 watts that gets as hot as 170 degrees C when run at only 16 watts, which is about 1/3 of the total power rating. Hot things like this have to be handled more carefully because they could end up ruining other parts nearby or even the circuit board if they are mounted on one.

 

Hi,

Yes that seems to work. The larger value resistor sets the limit on the current and the current is the same in both resistors so the max through both must be the max for the larger value resistor. Knowing this, the total power dissipation is the sum of the power dissipation in each with the max current flowing in each one. That totals 1500 watts.

You might note however that as a practical application this might not work in real life because if they really are resistors then the 20 ohm resistor would get super hot because it is run at the maximum power, which is probably based on a 20 or 25 degree C ambient anyway. The better choice would be to run it at 1/2 the rated power, which would then allow it to stay cooler.
At the max current level the 20 ohm resistor would get super hot while the 10 ohm resistor would stay cooler because it would be run at only 1/2 it's rated power.
The exception is if they are heater elements, which you might want to run at the normal operating current.
To give a quick example, i have a resistor that is rated for 50 watts that gets as hot as 170 degrees C when run at only 16 watts, which is about 1/3 of the total power rating. Hot things like this have to be handled more carefully because they could end up ruining other parts nearby or even the circuit board if they are mounted on one.

Thanks for explanation.

 

Resistor manufacturers seem to work diligently on making resistors that can survive intense heat. The problem with that is, they unsolder themselves! You can not trust a resistor to run at its rated wattage in a circuit board.

 

If we have two resistors R1=10Ω and R2=20Ω , the both have same maximal allowed power Pmax=1kW. Find the maximal power of the series of these two resistors?

I thought i could do it this way:

since P=I^2*R so if i have Pmax i can find Imax for both cases:

Imax1=sqrt(P1max/R1)=+-10A
Imax2=sqrt(P2max/R2)=+-5*sqrt(2)

since the Imax2 is lower than Imax1 it means that maximum current through the series is 5*sqrt(2), since Re=R1+R2=30Ω
Pmaxe=(Imax2)^2*Re=50*30=1,5kW.

Is this good approach, or im missing something?

That is a perfectly valid approach (though it would be nice if you tracked your units properly).

Another approach is to consider that, at a given current, the power goes linearly with the resistance. So a current that yields a certain power in one resistor will deliver half that power in a resistor half the size and twice that power in a resistor twice the size. This immediately tells you that the resistor that will be dissipating the most power will be R2 and that R1 will be delivering half of whatever R2 is dissipating. The means that

Ptot = P_R1 + P_R2
Ptot = 50% P_R2 + P_R2 = 150% P_R2
Ptotmax = 150% Pmax = 150% 1kW = 1.5 kW

 

Resistor manufacturers seem to work diligently on making resistors that can survive intense heat. The problem with that is, they unsolder themselves! You can not trust a resistor to run at its rated wattage in a circuit board.

Hi,

Yes good point. Some power resistors are rated up to 375 degrees C at 100 percent power. That's as hot as my soldering iron sometimes.
I just measured a large power resistor today and it was 233 degrees C at half power and still rising! I had to stop the test at that point because it was starting to melt the rubber on my clip lead alligator clip.