- #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 text book 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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