Max Duty Cycle for Undamaged Power Resistor (20V, 33ohm, 5W)

  • Thread starter peripatein
  • Start date
  • Tags
    Cycle
In summary, the conversation discussed finding the DC current and power dissipation for a resistor with a 20V source and a maximum power rating of 5 Watts. It was determined that the DC current running through the resistor was 0.6A and the power dissipation was 11.9W. The maximal DC current that could be run through the resistor without damage was found to be 0.39A. The conversation then turned to determining the maximal duty cycle allowed for the active load to remain undamaged, with the final equation being Pavg=(I*D)2*R, where I is the current through the resistor with 20 volts across it, and Iavg is the average current through the resistor due to the duty cycle.
  • #1
peripatein
880
0

Homework Statement


Assuming I have a 20V source on a “power resistor” which can withstand powers of up to 5 Watts on average before it excessively heats and damage begins. What is the DC current that runs through a resistor of 33ohm? What is its power dissipation? What is the maximal DC current I may run through it without damage, I5W? I am then asked to determine the maximal duty cycle allowed so the active load remains undamaged?

Homework Equations

The Attempt at a Solution


I found the DC current running through the resistor of 33ohm to be equal to 20/33=0.6A. The power equals I2R=11.9W. I5W=sqrt(5/33)=0.39A. Could the maximal duty cycle be found by simply dividing I5W by the first current, i.e. 0.39/0.6? If so, I am not quite sure why that is. Could someone kindly explain?
 
Physics news on Phys.org
  • #2
peripatein said:

Homework Statement


Assuming I have a 20V source on a “power resistor” which can withstand powers of up to 5 Watts on average before it excessively heats and damage begins. What is the DC current that runs through a resistor of 33ohm? What is its power dissipation? What is the maximal DC current I may run through it without damage, I5W? I am then asked to determine the maximal duty cycle allowed so the active load remains undamaged?

Homework Equations

The Attempt at a Solution


I found the DC current running through the resistor of 33ohm to be equal to 20/33=0.6A. The power equals I2R=11.9W. I5W=sqrt(5/33)=0.39A. Could the maximal duty cycle be found by simply dividing I5W by the first current, i.e. 0.39/0.6? If so, I am not quite sure why that is. Could someone kindly explain?
What would be the units of power/current? Do you think it could be a correct answer?
 
  • #3
It's 0.39A (not Watt!). I see no problem with the units, nor with dividing 0.39 by 0.6 in order to obtain a figure which has no units, such as duty cycle.
 
  • #4
P=I2*R

solve for I. Your resistor value is set (33 ohms). now you can find the current at any given power...
edit: oops, you already knew thatper your question, think about what you are doing.
you are taking your max current assuming it is allways on, and dividing it by the max current your part can take.

Iavg=I*D
Pavg=Iavg2*R
we know Iavg=I*D
so
Pavg=(I*D)2*R

you know what I will be based on your earlier equations, solve for D.
 
Last edited:
  • #5
Let's see whether I follow:
What's Iavg? Is it 20v/33ohm=0.6A?
And what's I? Is it I5W?
 
  • #6
peripatein said:
Let's see whether I follow:
What's Iavg? Is it 20v/33ohm=0.6A?
And what's I? Is it I5W?
sorry I was kind of scatterbrained there.

I is the current through the resistor with 20 volts across it.
Iavg is the average current through the resistor due to the duty cycle. so yes, by your notation Iavg=I5w
 
  • #7
Great, thank you!
 
  • #8
peripatein said:
It's 0.39A (not Watt!). I see no problem with the units, nor with dividing 0.39 by 0.6 in order to obtain a figure which has no units, such as duty cycle.
You are making my point. Units of amps are NOT a dimensionless ratio, which is what a duty cycle is.
 
  • #9
What I carried out was Amp/Amp!
 
  • #10
peripatein said:
What I carried out was Amp/Amp!
I was looking at your statement "dividing I5W by the first current" which is what I was alerting you to. Watts divided by amps is NOT dimensionless, it's volts.
 
  • #11
I5W is capital I (=current) and 5W as intended subscript, i.e. current for P=5W. Never mind.
 
  • #12
peripatein said:
I5W is capital I (=current) and 5W as intended subscript, i.e. current for P=5W. Never mind.
Ah ha. Clearly I was not paying close enough attention. Sorry about that.
 

1. What does "Max Duty Cycle" mean for an undamaged power resistor?

The Max Duty Cycle refers to the maximum amount of time that a power resistor can safely operate at its maximum power rating without overheating or causing damage.

2. How is the Max Duty Cycle determined for a power resistor?

The Max Duty Cycle is determined by the manufacturer through testing and analysis of the resistor's materials and design, as well as its ability to dissipate heat. It is typically expressed as a percentage of the total time that the resistor can safely operate at its maximum power rating.

3. What is the significance of the values 20V, 33ohm, and 5W in the context of Max Duty Cycle for an undamaged power resistor?

These values represent the voltage, resistance, and power rating of the power resistor. They are used to calculate the Max Duty Cycle by taking into account the resistor's ability to handle a certain amount of power without overheating.

4. Can the Max Duty Cycle be exceeded for an undamaged power resistor?

It is not recommended to exceed the Max Duty Cycle for an undamaged power resistor as it can lead to overheating and potential damage to the resistor. It is important to ensure that the power resistor is properly rated and used within its specified limits.

5. How does the Max Duty Cycle differ for damaged power resistors?

A damaged power resistor may have a reduced Max Duty Cycle due to a decrease in its ability to dissipate heat. This can occur if the resistor has been exposed to excessive heat or current, or if there is physical damage to the resistor. It is important to regularly inspect and replace damaged power resistors to ensure safe and efficient operation.

Similar threads

What Wattage Rating Is Needed for a 33Ohm Resistor in a 9V, 1A Circuit?
    • Electrical Engineering
3
Replies
78
Views
8K
Electrical A speed limiter for a Universal Motor in an old Lionel Train
    • DIY Projects
Replies
31
Views
6K
Find the Resistor Size for 160W Power Delivery: Kirchoff's Law Question"
    • Introductory Physics Homework Help
Replies
11
Views
1K

Hot Threads

Recent Insights

Back
Top