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

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Discussion Overview

The discussion revolves around determining the maximum duty cycle for a power resistor with a specified voltage, resistance, and power rating. Participants explore calculations related to current, power dissipation, and the implications of duty cycle on the resistor's operation without damage.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the DC current through a 33-ohm resistor with a 20V source to be 0.6A and the power dissipation to be 11.9W.
  • Another participant proposes that the maximum current without damage, denoted as I5W, is calculated as sqrt(5/33) resulting in 0.39A.
  • There is a suggestion to find the maximum duty cycle by dividing I5W by the initial current (0.39/0.6), though the reasoning behind this is questioned.
  • Some participants clarify the definitions of Iavg and I, with Iavg being the average current due to the duty cycle and equated to I5W.
  • Disagreement arises regarding the units involved in the calculations, particularly whether dividing currents yields a dimensionless duty cycle.
  • Clarifications are made about the notation used for I5W, indicating it represents current for a power of 5W.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the calculations and the interpretation of units, particularly regarding the duty cycle. No consensus is reached on the correct approach to calculating the maximum duty cycle.

Contextual Notes

Participants note potential confusion regarding the definitions of average current and maximum current, as well as the implications of unit conversions in the context of duty cycle calculations.

peripatein
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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?
 
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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?
 
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.
 
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:
Let's see whether I follow:
What's Iavg? Is it 20v/33ohm=0.6A?
And what's I? Is it I5W?
 
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
 
Great, thank you!
 
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.
 
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.
 

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