# Energy dissipated by resistor problem

## 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? 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?

#### Attachments

Doc Al
Mentor
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.

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.
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.

Doc Al
Mentor
But let's say they asked for the power generated by the resistor for the 5 second interval.
Don't worry about asking trivial questions! Those are the ones that drive you nuts. • 