A series RLC circuit has R = 420 , L = 1.45 H, C = 3.9 F. It is connected to an
ID: 1794638 • Letter: A
Question
A series RLC circuit has R = 420 , L = 1.45 H, C = 3.9 F. It is connected to an AC source with f = 60.0 Hz and Vmax = 150 V.
SOLVE IT
(A) Determine the inductive reactance, the capacitive reactance, and the impedance of the circuit.
Conceptualize The circuit of interest in this example is shown in the figure. The current in the combination of the resistor, inductor, and capacitor oscillates at a particular phase angle with respect to the applied voltage.
Categorize The circuit is a simple series RLC circuit, so we can use the approach discussed in this section.
Analyze
Find the angular frequency:
Use the following equation to find the inductive reactance:
Use the following equation to find the capacitive reactance:
Use the following equation to find the impedance:
(B) Find the maximum current in the circuit.
Use the following equation to find the maximum current:
(C) Find the phase angle between the current and voltage.
Use the the following equation to calculate the phase angle:
(D) Find the maximum voltage across each element.
Use the following equations to calculate the maximum voltages:
(E) What replacement value of L should an engineer analyzing the circuit choose such that the current leads the applied voltage by 30.0°? All other values in the circuit stay the same.
Solve the following equation for the inductive reactance:
XL = XC + R tan
Substitute expressions for inductive reactance and capacitive reactance into the following expression:
L =
+ R tan
Solve for L:
L =
+ R tan
Substitute the given values:
Finalize Because the capacitive reactance is larger than the inductive reactance, the circuit is more capacitive than inductive. In this case, the phase angle is negative, so the current leads the applied voltage. By applying the phase relationship to its instantaneous voltage, the instantaneous voltages across the three elements are:
vR = (143 V)sin(377t)
vL = (186 V)cos(377t)
vC = (231 V)cos(377t)
MASTER ITHINTS: GETTING STARTED | I'M STUCK!
Suppose the frequency is now increased to f = 93 Hz, and we want to keep the impedance unchanged.(a) What new resistance should we use to achieve this goal?
R =
(b) What is the phase angle (in degrees) between the current and the voltage now?
= °
(c) Find the maximum voltages across each element.
Explanation / Answer
initially,
R = 420 ohm
XL = 2 pi x 60 x 1.45 = 546.6 ohm
Xc = 1 / (2 pi f C ) = 680 ohm
Z = sqrt[ R^2 + (XL - Xc)^2] = 441 ohm
(A) Now f = 93 Hz
XL = 2 pi f L = 2 x pi x 93 x 1.45 = 847.3 ohm
Xc = 438.8 ohm
Z^2 = R^2 + (XL - Xc)^2 = 441^2
R^2 = 441^2 - (847.3 - 438.8)^2
R^2 = 441^2 - 408.5^2
R = 166 ohm
(b) theta = tan^-1 ((XL - XC)/R)
= 67.9 deg
(C) I = V / Z = 150 / 441 = 0.340 A
VR = 56.4 Volt
VL = 288 Volt
Vc = 149 Volt
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