Circuits: Average Power Absorbed by Elements

In summary: In general, the integrand will be a function of time (both v and i are). However, if v and i are in phase then i and v are proportional at all times, the power is the integral of vi and the power is constant. This is not true in general.In summary, the average power absorbed by a pure resistor, capacitor, and inductor when a sinusoidal voltage is applied is not always zero. It is zero only when the time interval of interest is a multiple of the period of the voltage. Otherwise, the power will vary over time and cannot be considered constant. This is because the instantaneous power is given by P(t)=i(t)v(t), and for sinusoidal voltage, the power is not constant unless
  • #1
Saladsamurai
3,020
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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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  • #2
Carrying out the procedure for a capacitor, I have the following:

Screenshot2011-01-22at13155PM.png


It seems to me that this is only zero when the t = t0 + nT where T is the period of sin(2wt) and n = 1,2,3 ...


Any thoughts? Is this correct that the average power is only zero over multiples of T or am I missing something?

Thanks!
 
  • #3
The text statement is incorrect.

The power certainly varies over time, which is why the exercise in finding average power is useful in the first place.

The instantaneous power is P(t)=i(t)v(t). For sinusoidal voltage it's evident that the power is not constant.

In order that the integral be invariant over any interval the integrand must be constant.
 
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What is the formula for calculating average power absorbed by an element in a circuit?

The formula for calculating average power absorbed by an element in a circuit is P = V * I * cos(θ), where P is power in watts, V is voltage in volts, I is current in amps, and θ is the phase angle between voltage and current.

How is average power absorbed by an element different from instantaneous power?

Average power absorbed by an element is the average rate at which energy is consumed by the element over a period of time. Instantaneous power, on the other hand, is the power consumed or produced by the element at a specific moment in time.

What is the unit of measurement for average power absorbed by an element?

The unit of measurement for average power absorbed by an element is watts (W).

Can average power absorbed by an element be negative?

Yes, average power absorbed by an element can be negative if the element is a source of energy, such as a battery or generator, and is supplying power to the circuit.

Why is it important to calculate average power absorbed by elements in a circuit?

Calculating average power absorbed by elements in a circuit is important because it helps determine the overall power consumption and efficiency of the circuit. It also allows for the proper sizing and selection of components to ensure the circuit operates safely and efficiently.

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