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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.
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.