Solve Series RLC Circuit for Capacitor Value

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Homework Help Overview

The discussion revolves around a series RLC circuit where the goal is to determine the value of a variable capacitor. The circuit parameters include a maximum voltage of 160 V, a frequency of 53 Hz, a 760 Ω resistor, and a 5.7 H inductor. The specific requirement is to find the capacitor value such that the voltage across it is out of phase with the applied voltage by 56 degrees.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the phase angle of the voltage across the capacitor and the current, questioning the interpretation of the phase angle and its implications on the calculations. There is an exploration of the equations related to impedance and phase relationships.

Discussion Status

Some participants have provided guidance on the interpretation of phase angles and the relationships between the voltage across the capacitor and the current. There is acknowledgment of potential errors in the angle used in calculations, and the discussion remains open with various interpretations being explored.

Contextual Notes

Participants note that the angle used in calculations may need reconsideration, particularly regarding the phase relationships in the circuit. There is also mention of a deadline for the assignment, indicating a time constraint for the original poster.

Idividebyzero
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1. Consider a series RLC circuit. The applied
voltage has a maximum value of 160 V and
oscillates at a frequency of 53 Hz. The circuit
contains a variable capacitor, a 760 Ω resistor,
and a 5.7 H inductor.
Determine the value of the capacitor such
that the voltage across the capacitor is out of
phase with the applied voltage by 56 degrees.

Answer in units of µF.




2. cos(phi)= R/Z
Z=SQRT(R^2 + (X_L-X_C)^2)
X_L= 2*pi*f*L
X_C=1/2*pi*f*C




3.first solved the equation cos(phi)=R/Z for the impedence Z. Z=R/cos(phi)

then proceeded to use the Z value in the second equation Z=SQRT(R^2 + (X_L-X_C)^2). squared both sides. then squared the given R. Subtracted R to the other side. (Z^2-R^2)= (X_L-X_c)2

square root both sides. then subracted X_l=2*Pi*f*l from the right to the left side. leaves a negative value on the left and the right so that negatives cancel, leaving a numerical value on the left and a X_c=1/2*pi*f*C on the right. inverted the left and the right. then divided the 2*pi*f on both sides. leaving c. the answer was incorrect.
 
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Keep in mind that you're looking for the relative phase of the voltage across the capacitor, not the phase of the current with respect to the voltage. The voltage across the capacitor will lag the current (by how much?).
 
the only thing that i can think of is the angle is wrong... 90-phi ?
 
The angle \phi that you've found is the angle by which the supply voltage leads the current in the circuit. The voltage on the capacitor will lag the current by 90°.

A simpler expression for \phi is
\phi = atan\left(\frac{X_L - X_C}{R}\right)
If you determine what \phi should be given the relationships between the relative angles of the voltage on the capacitor and the current, and the current and the voltage supply, you should be home-free.
 
thanks I've called it a night, going to tackle it again in the morning. assignments not due until friday and I've only got this problem left
 
gneill said:
The angle \phi that you've found is the angle by which the supply voltage leads the current in the circuit. The voltage on the capacitor will lag the current by 90°.

A simpler expression for \phi is
\phi = atan\left(\frac{X_L - X_C}{R}\right)
If you determine what \phi should be given the relationships between the relative angles of the voltage on the capacitor and the current, and the current and the voltage supply, you should be home-free.

This was it except the Phi term is acually 90-phi
 

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