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

[peripatein]

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?

 

[phinds]

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?

 

[peripatein]

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.

 

[donpacino]

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
P_avg=I_avg²*R
we know I_avg=I*D
so
P_avg=(I*D)²*R

you know what I will be based on your earlier equations, solve for D.

 

[peripatein]

Let's see whether I follow:
What's I_avg? Is it 20v/33ohm=0.6A?
And what's I? Is it I5W?

 

[donpacino]

Let's see whether I follow:
What's I_avg? 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.
I_avg is the average current through the resistor due to the duty cycle. so yes, by your notation I_avg=I_5w

 

[peripatein]

Great, thank you!

 

[phinds]

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.

 

[peripatein]

What I carried out was Amp/Amp!

 

[phinds]

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.

 

[peripatein]

I5W is capital I (=current) and 5W as intended subscript, i.e. current for P=5W. Never mind.

 

[phinds]

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.