Calculating RLC Circuit Power with Frequency: Solutions & Equations

In summary: V = Vo sin(ωt) and ω = 2*pi*f. Hope that was helpful.In summary, to calculate the power in an RLC circuit with an omega frequency power supply, write down an expression for power consumption in terms of I, V, omega, R, L or C in 't' period of time.
  • #1
Slepton
21
0

Homework Statement



How do one calculate the power in time of RLC circuit in parallel combination running an AC current ? It has to be in terms of frequency ([tex]\omega[/tex] = 2 [tex]\pi[/tex] f )

Homework Equations



[tex]\omega[/tex] = 1/ [tex]\sqrt{LC}[/tex]

The Attempt at a Solution



I started off with the omega but could not reach all the way to an expression for power in terms of omega.

Thanks in advance
 
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  • #2
Slepton said:

Homework Statement



How do one calculate the power in time of RLC circuit in parallel combination running an AC current ? It has to be in terms of frequency ([tex]\omega[/tex] = 2 [tex]\pi[/tex] f )

Homework Equations



[tex]\omega[/tex] = 1/ [tex]\sqrt{LC}[/tex]

The Attempt at a Solution



I started off with the omega but could not reach all the way to an expression for power in terms of omega.

Thanks in advance

Welcome to the PF. You need to define the problem a bit more. An RLC circuit does not dissipate power unless driven by some source, and that source will have some source impedance. Otherwise, do you mean that the capacitor is charged up to some voltage at time t=0, and then the charging source is removed, and you want to plot the power dissipated versus time?
 
  • #3
sorry, I was vague. So, omega is the frequency of the power supply. I and V are the total current and voltages of the circuit. L, R, C are the inductance, resistance and capacitance. I have to find power consumption due to the parallel LCR circuit for a time 't'. I am told to express the power in I, V, R, omega, L or C in time 't'. If I and V are the total current and voltage, wouldn't power just be IV. But since it's AC circuit, I and V fluctuate.
 
  • #4
Slepton said:
sorry, I was vague. So, omega is the frequency of the power supply. I and V are the total current and voltages of the circuit. L, R, C are the inductance, resistance and capacitance. I have to find power consumption due to the parallel LCR circuit for a time 't'. I am told to express the power in I, V, R, omega, L or C in time 't'. If I and V are the total current and voltage, wouldn't power just be IV. But since it's AC circuit, I and V fluctuate.

Still not enough information. Please post the exact text of the question, along with any diagrams. And post your work towards a solution. As you have read at the PF Rules link at the top of this page, we cannot help until you post your work.
 
  • #5
The question was made by my professor. So, that's all the info. it gives us. It reads exactly like this:

If you are using a parallel RLC circuit with omega as the frequency of the power supply, I and V as the total current and voltage of the circuit, write down an expression for the power consumption in terms of I, V, omega, R, L or C in 't' period of time.

What I did so far is wrote down power as P= V^2/Z = V_0 exp (2*(j*omega*t -j*phi))*(1/R + j(1/X_c - 1/X_L)). I got confused from here since, I am asked for power in time 't' and power already is work per unit time. so, does it mean that i have to again take derivative of P with respect to time?
 
  • #6
't' period of time.
What is the relation between 't' period of time and ω?
 
  • #7
If t was time period then t = 2*pi/omega, but from the question it sounds like t could be any time.
 
  • #8
I'm the retired high school teacher hoping to learn something here, too.
Slepton, we teachers are not always perfect and your teacher has a mistake in the question - it is not correct to say "the power in time t". Rather you must say the "energy in time t" or the "power at time t". No doubt it is the latter we are looking for.

Now, I don't see how or why you got that complicated expression. To my high school mind, it should be just
P = V^2/Z = Vo^2*e^(2jwt) all divided by R + j(Xc - XL).
 
  • #9
In parallel RLC circuit, RL are in series and C is in parallel to it. In this circuit the power dissipation is in the resistance only. Hence P = [ V/( R + jωL )]^2*R where V = Vosin(ωt)
 

FAQ: Calculating RLC Circuit Power with Frequency: Solutions & Equations

What is an RLC circuit?

An RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C). These components are connected in series or in parallel and their values determine the behavior of the circuit.

How do you calculate the power in an RLC circuit?

The power in an RLC circuit can be calculated using the equation P = VI, where P is power, V is voltage, and I is current. In an RLC circuit, the voltage and current are not in phase with each other, so the power must be calculated using the complex power formula: P = Re(VI*), where Re is the real part and VI* is the complex conjugate of the voltage and current.

What is the relationship between frequency and power in an RLC circuit?

In an RLC circuit, the power is directly proportional to the frequency. This means that as the frequency increases, the power also increases. This relationship can be represented by the equation P = VI*cos(θ), where θ is the phase angle between the voltage and current.

How do you solve for power in an RLC circuit using complex numbers?

To solve for power in an RLC circuit using complex numbers, you can use the complex power formula: P = Re(VI*), where VI* is the complex conjugate of the voltage and current. This formula takes into account the phase difference between the voltage and current, which is important in RLC circuits.

What is the significance of power factor in an RLC circuit?

Power factor is a measure of how efficiently a circuit uses electrical power. In an RLC circuit, the power factor is affected by the phase difference between the voltage and current. A power factor of 1 means that the voltage and current are in phase and the circuit is using all of the available power. A power factor less than 1 means that there is a phase difference and the circuit is not using all of the available power.

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