How Do You Calculate Impedance in a Series RLC Circuit?

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SUMMARY

The impedance of a series RLC circuit consisting of a 1170-ohm resistor, a 1.46 μF capacitor, and a 2.7 H inductor connected to a 350 Hz AC source with a maximum voltage of 230 V is calculated to be approximately 6.05179 kiloohms. The calculations utilize the formulas for inductive reactance (XL = 2πfL) and capacitive reactance (XC = 1 / (2πfC)). The values obtained for XL and XC are 5937.61 ohms and 0.000245 ohms, respectively. The final impedance is derived from the equation Z = sqrt(R^2 + (XL - XC)^2).

PREREQUISITES
  • Understanding of series RLC circuits
  • Familiarity with AC circuit analysis
  • Knowledge of reactance calculations (XL and XC)
  • Proficiency in using the impedance formula
NEXT STEPS
  • Study the effects of frequency on impedance in RLC circuits
  • Learn about resonance in series RLC circuits
  • Explore phasor diagrams for AC circuit analysis
  • Investigate the impact of component tolerances on impedance calculations
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Electrical engineering students, educators, and professionals involved in circuit design and analysis, particularly those focusing on AC circuit behavior and impedance calculations.

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Homework Statement


A 1170 resistor, a 1.46 μF capacitor, and
a 2.7 H inductor are connected in series across
a 350 Hz AC source for which the maximum
voltage is 230 V.
Calculate the impedance of the circuit.


Homework Equations



Z=sqrt(R^2+(XL-XC)^20

XL=2*pi*f*L

XC=1 / (2*pi*f*C)

The Attempt at a Solution



XL= 2*pi*(350Hz)*(2.7H) = 5937.61

XC= 1 / (2*pi*350Hz* 1.46e-6 F) = .000245

Z= sqrt( 1170^2 + (5937.61 - 2.54e-4)^2) = 6051.79 ohms => 6.05179 kiloohms
 
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Your Xc calculation looks wrong... should be up closer to 100 Ohms or so...
 


awesome! thanks!
 

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