Maximum energy stored in a capacitor? RLC circuit

Click For Summary
SUMMARY

The discussion centers on calculating the maximum energy stored in a capacitor within a series RLC circuit, specifically when the current leads the generator voltage by an angle φ = 30°. The circuit includes an inductor of L = 400 mH, a resistor of R = 50 Ω, and operates at a generator frequency of ω = 100 rad/s with a maximum emf of 10 V. The capacitance is determined to be 145e-6 F, and the energy stored in the capacitor at time t = π/2ω is expressed as UC = UC,max sin²φ.

PREREQUISITES
  • Understanding of RLC circuit dynamics
  • Knowledge of phasor diagrams
  • Familiarity with the formula for energy stored in a capacitor: U = 1/2 * C * V²
  • Basic concepts of impedance in AC circuits
NEXT STEPS
  • Study the derivation of energy formulas in RLC circuits
  • Learn about phasor analysis in AC circuit theory
  • Explore the impact of phase angles on circuit behavior
  • Investigate the relationship between capacitance and impedance in AC circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing RLC circuits in AC applications.

Luke Cohen
Messages
30
Reaction score
1

Homework Statement


The current in a series RLC circuit leads the generator voltage by φ = 30°. The circuit, containing an inductor L = 400 mH and a resistor R = 50 Ω, is driven by a generator operating at ω = 100 rad/s with a maximum emf of 10 V. The capacitance is unknown.

The first question regarding this prompt is to find the capacitance, which is 145e-6F.

This is the question I am stumped on:

In terms of the maximum energy UC,max stored in the capacitor and the angle φ by which the current leads the generator voltage, the energy UC stored in the capacitor when the time is t = π/2ω is:

The correct answer is UC = UC,max sin^2φ

Homework Equations

The Attempt at a Solution


I've drawn my phasor diagrams, but I don't understand the correct answer. Is the voltage across the capacitor the component of the voltage across the generator pointing in the direction of X_c, the capacitor's impedance? If so, I get how a sin^2(phi) term would be used, because that would be the V^2 part of U = 1/2*cV^2, but I don't see how that would give me UC,max *sin^2(phi).
I will appreciate any help!
 
Physics news on Phys.org
This is just a normal box series RLC circuit with a battery, resistor, capacitor, and inductor in series. Sorry for not including!
 
Y
Luke Cohen said:

Homework Statement


The current in a series RLC circuit leads the generator voltage by φ = 30°. The circuit, containing an inductor L = 400 mH and a resistor R = 50 Ω, is driven by a generator operating at ω = 100 rad/s with a maximum emf of 10 V. The capacitance is unknown.

The first question regarding this prompt is to find the capacitance, which is 145e-6F.

This is the question I am stumped on:

In terms of the maximum energy UC,max stored in the capacitor and the angle φ by which the current leads the generator voltage, the energy UC stored in the capacitor when the time is t = π/2ω is:

The correct answer is UC = UC,max sin^2φ

Homework Equations

The Attempt at a Solution


I've drawn my phasor diagrams, but I don't understand the correct answer. Is the voltage across the capacitor the component of the voltage across the generator pointing in the direction of X_c, the capacitor's impedance? If so, I get how a sin^2(phi) term would be used, because that would be the V^2 part of U = 1/2*cV^2, but I don't see how that would give me UC,max *sin^2(phi).
I will appreciate any help!
You have to work with time dependence of the current and voltages. Assuming the generator voltage is Vg=Vg0sin(ωt),
what is the time dependence of the current and the capacitor voltage?
 
Luke Cohen said:
This is just a normal box series RLC circuit with a battery, resistor, capacitor, and inductor in series. Sorry for not including!

There is no escaping this: you will need to show, and then use, the phasor diagram of current and voltages.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 15 ·
Replies
15
Views
912
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K