Solve Series RLC Circuit for Capacitor Value

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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 [itex]\phi[/itex] 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 [itex]\phi[/itex] is
[tex]\phi = atan\left(\frac{X_L - X_C}{R}\right)[/tex]
If you determine what [itex]\phi[/itex] 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 [itex]\phi[/itex] 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 [itex]\phi[/itex] is
[tex]\phi = atan\left(\frac{X_L - X_C}{R}\right)[/tex]
If you determine what [itex]\phi[/itex] 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