- #1
Saladsamurai
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Homework Statement
Let [itex]v(t) = V_{max}\cos(\omega t )[/itex] be applied to (a) a pure resistor, (b) a pure capacitor (with zero initial capacitor voltage) and (c) a pure inductor (with zero initial inductor voltage). Find the average power absorbed by each element.
Homework Equations
[tex]P_{avg}(\tau) = \frac{1}{t}\int_0^t i(\tau)v(\tau)\,d\tau[/tex]
Ohms law.
The Attempt at a Solution
Please skip ahead to the capacitor and inductor part of problem.Resistor
combining Ohm's Law and the average power equation, we have
[tex]P_{avg,R} = \frac{1}{t}\int_0^t \frac{v(\tau)}{R}v(\tau)\,d\tau[/tex]
[tex] = \frac{1}{Rt}\int_0^t \left [V_{max}\cos(\omega \tau )\right ] ^2 \,d\tau[/tex]
Using the trig identity: [itex]2\cos(x)\cos(y) = \cos(x - y) + \cos(x+y)[/itex]
[tex]P_{avg,R} = \frac{V_{max}^2}{Rt}\int_0^t \cos(\omega \tau )\cos(\omega \tau )d\tau[/tex]
[tex] = \frac{V_{max}^2}{2Rt}\int_0^t \left \{1 - \cos(2\omega\tau)\right \} \,d\tau[/tex]
[tex]\Rightarrow P_{avg,R} = \frac{V_{max}^2}{2Rt}\left \{\tau - \frac{1}{2\omega}\sin(2\omega\tau)\right \}_0^t[/tex]
Hmmm ... I guess this does work. I thought that the units weren't going to check out, but I forgot that 1/omega gets me my time dimension back. Oh well. I guess I answered this part ok.
For the capacitor and inductor I am assuming I will go through a similar procedure, however, my textbook makes a claim in the chapter that I disagree with. It says that the average power of a capacitor and inductor whose voltage and current vary sinusoidally can be shown to be zero. Is this ALWAYS true? Or only when the time interval of interest is some multiple of the period?
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