Solving Series RLC Circuit: Vrms Across Capacitor & Inductor is Zero

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SUMMARY

The discussion focuses on determining the appropriate capacitance (C) in a series RLC circuit to achieve zero Vrms across the capacitor and inductor. Given the parameters R = 10 Ω, L = 0.1 H, and a power source of V = 5sin(500t), the resonance frequency equation ω² = 1/LC is utilized. By substituting ω = 500 and L = 0.1 H, the calculated capacitance is C = 0.00004 F. The relationship between the voltages across the inductor and capacitor is clarified, emphasizing that they are 180° out of phase, confirming resonance conditions.

PREREQUISITES
  • Understanding of series RLC circuits
  • Knowledge of resonance frequency equations
  • Familiarity with phasor representation in AC circuits
  • Basic concepts of impedance (XL and XC)
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  • Learn about phasor analysis in AC circuit theory
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Electrical engineering students, circuit designers, and anyone studying AC circuit analysis and resonance in RLC circuits.

reddawg
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Homework Statement


Given a series RLC circuit:

R = 10 Ω
L = 0.1 H
C = 2 μF
Power Source: V = 5sin(500t)

If you could change the capacitance what C would you select so Vrms across the capacitor and inductor is zero?

Homework Equations


ω2 = 1/LC

The Attempt at a Solution


My instructor's solutions say to use above equation for resonance frequency to input ω=500 (given from V) and
L = 0.1 H and then solve for C, yielding 0.00004 F.

This makes no sense, could someone explain it to me?
 
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reddawg said:
This makes no sense, could someone explain it to me?
Remember: The voltage across the inductance is 180° out of phase with respect to the voltage across the capacitor (jωL vs. 1/[jωC]).
 
Svein said:
Remember: The voltage across the inductance is 180° out of phase with respect to the voltage across the capacitor (jωL vs. 1/[jωC]).
Ok, I think I get it.

Given a phaser representation, tan(φ) = (XL - XC) / R

So tan(180) = 0 ⇒ XL = XC which is resonance.
 

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