What Are the Correct Calculations for Reactance in a Parallel RLC Circuit?

In summary, the conversation is about a problem with a parallel RLC circuit and finding the reactance for a given frequency. The individual has made some calculations but needs help as their answers were marked wrong. The correct approach was used but there may be an error in the calculations.
  • #1
DethRose
101
0
Hey

im in first year electronics and have a problem with a parallel rlc circuit

im trying to find the reactance (Xc and Xl) for a circuit that has a 500 ohm resister and a .01 microfarad capacitor in parallel with a 500 ohm resistor and a .5 mH inductor. The frequency of the circuit is 50 KhZ

here are my calculations

xc= 1/(2)(pie)(50khz)(.01microfarads)
= 0.32

xl= (2)(pie)(50khz)(.0005H)
=0.157

but i had them both marked wrong...help with this would be very much appreciated
 
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  • #2
DethRose said:
im trying to find the reactance (Xc and Xl) for a circuit that has a 500 ohm resister and a .01 microfarad capacitor in parallel with a 500 ohm resistor and a .5 mH inductor. The frequency of the circuit is 50 KhZ

here are my calculations

xc= 1/(2)(pie)(50khz)(.01microfarads)
= 0.32

xl= (2)(pie)(50khz)(.0005H)
=0.157
You have used the right approach to work out the capacitive reactance and inductive reactance but you have to check your math. You are out by a factor of 1000 in each case.

To find the total reactance you have to subtract the capacitive from the inductive reactance.

AM
 
  • #3


Hi there,

Thank you for sharing your question about a parallel RLC circuit. It seems like you have made some good progress in your calculations. However, there are a few things to consider:

1. The units for frequency should be in Hertz (Hz), not kilohertz (kHz). So, the frequency in your calculations should be 50,000 Hz instead of 50 kHz.

2. The formula for capacitance (C) is C = 1/(2πfXc), where f is the frequency in Hz and Xc is the capacitive reactance in ohms. So, your calculation for Xc should be:

Xc = 1/(2π*50,000*0.01*10^-6) = 318.3 ohms

3. Similarly, the formula for inductance (L) is L = Xl/(2πf), where f is the frequency in Hz and Xl is the inductive reactance in ohms. So, your calculation for Xl should be:

Xl = (2π*50,000*0.5*10^-3) = 157.1 ohms

I hope this helps clarify your calculations. Remember to always pay attention to the units and formulas when solving problems in electronics. Good luck with your studies!
 

Related to What Are the Correct Calculations for Reactance in a Parallel RLC Circuit?

1. What is a series RLC circuit?

A series RLC circuit is an electrical circuit that consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series. This means that the components are connected one after the other, and the same current flows through each component.

2. How does a series RLC circuit behave?

The behavior of a series RLC circuit depends on the values of the components and the frequency of the applied voltage. At low frequencies, the inductor behaves like a short circuit, and the capacitor behaves like an open circuit, resulting in a high impedance. At high frequencies, the opposite is true, and the impedance is low. At a specific frequency (known as the resonant frequency), the inductive and capacitive reactances cancel each other out, resulting in a purely resistive circuit.

3. What is the formula for calculating the impedance of a series RLC circuit?

The formula for calculating the impedance of a series RLC circuit is Z = √(R² + (XL - XC)²), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. This formula takes into account the effects of both the inductor and capacitor on the overall impedance of the circuit.

4. How do you analyze a series RLC circuit?

To analyze a series RLC circuit, you can use Kirchhoff's laws and Ohm's law to calculate the current and voltage at different points in the circuit. You can also use the formula for impedance to determine the behavior of the circuit at different frequencies. Additionally, you can use a circuit simulator or perform calculations using a computer program.

5. What are some practical applications of series RLC circuits?

Series RLC circuits have various practical applications, such as in tuning circuits (e.g., radio receivers), signal filters, and power factor correction. They are also used in electronic devices that require precise timing, such as oscillators and timers. Additionally, series RLC circuits are used in electrical power systems to regulate voltage and current and protect against overloads and short circuits.

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