Question
A series RLC circuit consists of
a 75Ω resistor, a 5.0 μf capacitor
and a 75mH inductor. They are
connected across a generator of
frequency 250Hz with a rms
voltage of 12V. Determine the
impedance of the circuit.
A. 127.32 Ω
B. 117.8 Ω
C. 84.51Ω
D. 74.39Ω
E. 75.6Ω
a 75Ω resistor, a 5.0 μf capacitor
and a 75mH inductor. They are
connected across a generator of
frequency 250Hz with a rms
voltage of 12V. Determine the
impedance of the circuit.
A. 127.32 Ω
B. 117.8 Ω
C. 84.51Ω
D. 74.39Ω
E. 75.6Ω
Answers
GPT 3.5
To find the impedance of the series RLC circuit, we need to calculate the total resistance (R), total inductive reactance (XL), and total capacitive reactance (XC), and then combine them using the impedance formula:
Impedance (Z) = √(R^2 + (XL - XC)^2)
Given:
R = 75 Ω
C = 5.0 μF
L = 75 mH
f = 250 Hz
Vrms = 12 V
First, let's calculate XL and XC using the formulas:
XL = 2πfL
XC = 1/(2πfC)
XL = 2π(250)(75 * 10^(-3)) = 0.942 Ω
XC = 1/(2π(250)(5 * 10^(-6))) = 127.323 Ω (rounded to 3 decimal places)
Next, let's calculate the impedance:
Z = √(R^2 + (XL - XC)^2)
Z = √((75)^2 + (0.942 - 127.323)^2)
Z = √(5625 + (-126.381)^2)
Z = √(5625 + 15982.854761)
Z = √(21607.854761)
Z = 146.951 Ω (rounded to 3 decimal places)
Therefore, the impedance of the circuit is approximately 146.951 Ω, which is not one of the given answer choices.
Impedance (Z) = √(R^2 + (XL - XC)^2)
Given:
R = 75 Ω
C = 5.0 μF
L = 75 mH
f = 250 Hz
Vrms = 12 V
First, let's calculate XL and XC using the formulas:
XL = 2πfL
XC = 1/(2πfC)
XL = 2π(250)(75 * 10^(-3)) = 0.942 Ω
XC = 1/(2π(250)(5 * 10^(-6))) = 127.323 Ω (rounded to 3 decimal places)
Next, let's calculate the impedance:
Z = √(R^2 + (XL - XC)^2)
Z = √((75)^2 + (0.942 - 127.323)^2)
Z = √(5625 + (-126.381)^2)
Z = √(5625 + 15982.854761)
Z = √(21607.854761)
Z = 146.951 Ω (rounded to 3 decimal places)
Therefore, the impedance of the circuit is approximately 146.951 Ω, which is not one of the given answer choices.
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