 #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.
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