Average power in series LC circuit

In summary, the conversation discusses a series LC circuit with a resonant frequency and a voltage source. The voltage amplitude of the capacitor is 1V and the task is to find the average power in the circuit. It is determined that the power is zero if there is no active resistance, but this is not a realistic scenario. The conversation suggests considering all possibilities, including the presence of a non-zero resistance, and providing alternative ways to calculate the power loss in order to demonstrate competency in the topic.
  • #1
Rugile
79
1

Homework Statement


A voltage source is connected to a series LC circuit. The frequency of the source is resonant. The voltage amplitude of capacitor is 1V. Find the average power in the circuit.

Homework Equations

The Attempt at a Solution


I realize that if there is no active resistance the power is 0. I tried assuming that there is active resistance of the voltage source, but it leads me nowhere since there are so many unknowns left: [itex]I_{max} = \frac{U_{max}}{R}[/itex]
$$P=\frac{U_{max}I_{max} }{2}$$
$$U_{maxC} = \frac{I_{max}}{\omega C} = \frac{U_{max}}{R \omega C} = \frac{U_{max} \sqrt{LC}}{RC}$$
I really don't know how to work around this... Or should I just assume that there is no active resistance? Any help appreciated!
 
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  • #2
Leave it at that: average power is zero.
For ANY finite input voltage V it is impossible for a pure L-C circuit to have 1V across the capacitor (or inductor for that matter). For ANY finite input voltage the voltage across either component would be infinite, as would be the current.
Of course, this circuit is non-realizable.
If you stuck a resistance R in series with the L-C section, the power would simply be (V^2)/R, again getting you nowhere.
 
  • #3
As the maximum voltage of the capacitor is finite, there is some resistance in the circuit. Assume that t the generator voltage and L, C are given. Determine the average power in terms of them.
 
  • #4
Rugile said:
I really don't know how to work around this... Or should I just assume that there is no active resistance? Any help appreciated!
In situations like this, try to cover all bases. You've considered that R may be zero. So if R is non-zero write the power loss in a couple of ways, e.g., (wC)2.R, V2/R, while explaining these terms.

By making the best of a bad situation (a badly contrived question) you at least demonstrate competency in the topic, and that is basically what any marker is looking for so he/she can justify awarding you marks. Make it easy for the marker---show what you know!
 
  • #5


I would approach this problem by first understanding the basic principles of a series LC circuit. In this type of circuit, the capacitor and inductor are connected in series, creating a resonant frequency at which the voltage across the capacitor and the current through the inductor are in phase. This means that the power in the circuit is purely reactive and there is no active power dissipated.

Therefore, in this scenario, the average power in the circuit would indeed be 0. This is because the average power is calculated by taking the time average of the instantaneous power, and since the power is constantly oscillating between positive and negative values, the average will always be 0.

In conclusion, if the frequency of the source is resonant and there is no active resistance, the average power in the series LC circuit will be 0. If there is active resistance, the average power will still be 0 since the power is purely reactive in this type of circuit.
 

What is an average power in series LC circuit?

An average power in series LC circuit refers to the amount of power that is dissipated or consumed by the circuit on average over a period of time. It is calculated by taking the average of the instantaneous power values over one cycle of the circuit.

How is average power calculated in a series LC circuit?

The average power in a series LC circuit can be calculated by using the formula P = Vrms x Irms x cos(θ), where Vrms is the root mean square (RMS) voltage, Irms is the RMS current, and cos(θ) is the power factor.

What is the significance of average power in series LC circuits?

The average power in series LC circuits is important because it helps to determine the efficiency of the circuit. It also helps to determine the amount of power that is being consumed or dissipated, which can affect the overall performance and lifespan of the circuit.

How does the value of the capacitor and inductor affect the average power in a series LC circuit?

The value of the capacitor and inductor in a series LC circuit can affect the average power in different ways. For example, a larger capacitor can lead to a decrease in average power, while a larger inductor can lead to an increase in average power.

What are some practical applications of series LC circuits and their average power?

Series LC circuits have many practical applications, including in radio frequency (RF) filters, sensors, and oscillators. Average power is important in these applications as it helps to determine the efficiency and performance of the circuit.

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