- #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?

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.