Energy dissipated by resistor problem

  • #1

Homework Statement


A 10 Ω resistor carries a current that varies as a function of time as shown. How much energy has been dissipated by the resistor after 5 s?
upload_2018-7-26_11-7-37.png

a. 40 J
b. 50 J
c. 80 J
d. 120 J

Homework Equations


I=dq/dt → dq=Idt → Q=∫Idt
P=∇E/∇t and V=IR (Ohm's Law) so P=IV ⇒ P=I^2R

The Attempt at a Solution


My approach[/B]
For the first 5 seconds, the area under the curve is 6C, which is the total charge that was transferred through the resistor. Thus, the average current I=6C/5s=6/5 A
Since P=I^2R and P=E/∇t ⇒ E=I^2R∇t=(6/5A)^2(10Ω)(5s)= 72J
Clearly this is none of the answer choices, and I don't know where I went wrong.

Book's Approach
The book claims that D is correct, and states:
Since the current was 2A over 3s, the total energy dissipated is
E=(2A)^2(10Ω)(3s)=120J

I genuinely don't understand the book's approach. Power is by definition, the rate of energy transfer (in this case transformation to a different form) with respect to time. Here, the power MUST be computed over the 5 seconds in question, regardless of whether there was 0 current passing through for 2 seconds.
Ultimately, why did the book use 3 seconds as ∇t? Is this allowed, and why is my approach flawed?
Thank you all in advance.
 

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Answers and Replies

  • #2
Doc Al
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For the first 5 seconds, the area under the curve is 6C, which is the total charge that was transferred through the resistor. Thus, the average current I=6C/5s=6/5 A
Since P=I^2R and P=E/∇t ⇒ E=I^2R∇t=(6/5A)^2(10Ω)(5s)= 72J
Clearly this is none of the answer choices, and I don't know where I went wrong.
You found the average current. But since power is proportional to current squared, to use your approach you'd need the average value of current squared. (Which is not equal to the square of the average current.)

Here, the power MUST be computed over the 5 seconds in question, regardless of whether there was 0 current passing through for 2 seconds.
They did compute the energy over all 5 seconds. Since there was no energy used during the time the current was zero, you can skip that segment.
 
  • #3
You found the average current. But since power is proportional to current squared, to use your approach you'd need the average value of current squared. (Which is not equal to the square of the average current.)


They did compute the energy over all 5 seconds. Since there was no energy used during the time the current was zero, you can skip that segment.
Thanks Doc. I am starting to understand why their approach works for computing energy. But let's say they asked for the power generated by the resistor for the 5 second interval.
Would the answer then be
P=120J/3s= 40W ?
I.e., can we again neglect that 2 second segment where no energy was generated? And if so, why?
Btw I'm sorry if I'm asking trivial questions, I've never been good at physics lol.
 
  • #4
Doc Al
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But let's say they asked for the power generated by the resistor for the 5 second interval.
Would the answer then be
P=120J/3s= 40W ?
If they asked for the average power over that interval, then the answer would be 120J/5s = 24 W. (That's what I would say.) You cannot neglect any part of the interval if you want the average power over it.

Don't worry about asking trivial questions! Those are the ones that drive you nuts. :smile:
 
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