How Do You Calculate R and C for a Series Impedance of 500∠-36.1° at 10kHz?

In summary: So with two equations and two unknowns, you should be able to solve for R and C. In summary, the problem is to compute values for R and C such that the total series impedance is 500\angle(-36.1o) when the frequency is 10kHz. This can be solved using two equations, one for the real component and one for the imaginary component of the impedance. With two equations and two unknowns, R and C can be determined.
  • #1
zealeth
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Homework Statement



Compute values for R and C such that the total series impedance Z=500[itex]\angle[/itex](-36.1o) when f=10kHz.

F5llcWq.png


Homework Equations



ω=2[itex]\pi[/itex]f
KVL
ZC = -j*(1/(ωC)) = 1/(jωC) where j=[itex]\sqrt{-1}[/itex]

The Attempt at a Solution



Seems like a pretty straightforward problem, I seem to be missing an equation somewhere but not sure what else I could use here.

ω=2*[itex]\pi[/itex]*10*103 = 62832 rad/s
Zeq=500[itex]\angle[/itex](-36.1o) = R - j/(ωC) = R - j/(62832*C)

And here I am left with 1 equation and 2 variables, any ideas on what else I could use to solve this problem?
 
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  • #2
zealeth said:

Homework Statement



Compute values for R and C such that the total series impedance Z=500[itex]\angle[/itex](-36.1o) when f=10kHz.

F5llcWq.png


Homework Equations



ω=2[itex]\pi[/itex]f
KVL
ZC = -j*(1/(ωC)) = 1/(jωC) where j=[itex]\sqrt{-1}[/itex]

The Attempt at a Solution



Seems like a pretty straightforward problem, I seem to be missing an equation somewhere but not sure what else I could use here.

ω=2*[itex]\pi[/itex]*10*103 = 62832 rad/s
Zeq=500[itex]\angle[/itex](-36.1o) = R - j/(ωC) = R - j/(62832*C)

And here I am left with 1 equation and 2 variables, any ideas on what else I could use to solve this problem?

Actually you're left with two equations, each in one variable. Note that the impedance has two terms, one real and one imaginary. The real and imaginary components are separate. For example, the real part of the impedance (the resistance) is independent of frequency.
 
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Related to How Do You Calculate R and C for a Series Impedance of 500∠-36.1° at 10kHz?

1. What is an R-C network?

An R-C network is a type of electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series. It is used to analyze and understand the behavior of alternating current (AC) circuits.

2. How does an R-C network work?

In an R-C network, the resistor and capacitor work together to control the flow of AC current. The resistor limits the current flow, while the capacitor stores and releases electrical energy. This interaction between the two components creates a time delay in the circuit, which affects the overall behavior of the AC current.

3. What is the purpose of R-C network analysis?

R-C network analysis is used to calculate and understand the characteristics of an AC circuit, such as voltage, current, and phase angle. It is an essential tool for designing and troubleshooting electronic circuits.

4. What are the key parameters in R-C network analysis?

The key parameters in R-C network analysis include resistance (R), capacitance (C), frequency (f), and phase angle (φ). These parameters are used to calculate important values such as impedance, reactance, and time constant.

5. How is R-C network analysis performed?

R-C network analysis involves using mathematical formulas, such as Ohm's law and Kirchhoff's laws, to calculate the various parameters of the circuit. These calculations can be done by hand or by using specialized software, such as SPICE. The results of the analysis can then be used to design and optimize the circuit for specific applications.

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