Solving circuit using complex numbers

In summary, the current in the circuit is I=\frac{U}{R+Z} or when it is arranged I=\frac{iU_0(1-CL\omega^2)}{iL\omega (2-CL\omega^2)}.
  • #1
masterjoda
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Homework Statement


Find the current in them in this circuit, if we know [itex]R=X_L, X_C[/itex] and [itex]u=5sin(314t)[/itex]


The Attempt at a Solution


First , [itex]5=U_0, 314=\omega[/itex] and voltage we can write as [itex]u=U_0cos(\omega t + \frac{\pi}{2})[/itex] and [itex]u=U_0 e^{i\frac{\pi}{2}}=iU_0[/itex]. [itex]U[/itex] is the voltage at the source [itex]U_1[/itex] in the branch and [itex]U_2[/itex] at the resistor. Now [itex]U=U_1+U_2[/itex] or [itex]U=IR+IZ[/itex] where [itex]\frac{1}{Z}=\frac{1}{iL\omega}-\frac{C\omega}{i}[/itex] or [itex]Z=\frac{iL\omega}{1-CL\omega^2}[/itex]
Now the current is [itex]I=\frac{U}{R+Z}[/itex] or when it is arranged [itex]I=\frac{iU_0(1-CL\omega^2)}{iL\omega (2-CL\omega^2)}[/itex]. Now I don't know how to complete the calculation to reduce it to the form like this [itex]I_0sin(\omega t + \theta)[/itex].
 

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  • #2
If it is true that R = XL,XC, then XL = XC. What does that tell you about the LC tank circuit?
 
  • #3
No just [itex]R=X_L[/itex].
 
  • #4
masterjoda said:

Homework Statement


Find the current in them in this circuit, if we know [itex]R=X_L, X_C[/itex] and [itex]u=5sin(314t)[/itex]


The Attempt at a Solution


First , [itex]5=U_0, 314=\omega[/itex] and voltage we can write as [itex]u=U_0cos(\omega t + \frac{\pi}{2})[/itex] and [itex]u=U_0 e^{i\frac{\pi}{2}}=iU_0[/itex]. [itex]U[/itex] is the voltage at the source [itex]U_1[/itex] in the branch and [itex]U_2[/itex] at the resistor. Now [itex]U=U_1+U_2[/itex] or [itex]U=IR+IZ[/itex] where [itex]\frac{1}{Z}=\frac{1}{iL\omega}-\frac{C\omega}{i}[/itex] or [itex]Z=\frac{iL\omega}{1-CL\omega^2}[/itex]
Now the current is [itex]I=\frac{U}{R+Z}[/itex] or when it is arranged [itex]I=\frac{iU_0(1-CL\omega^2)}{iL\omega (2-CL\omega^2)}[/itex]. Now I don't know how to complete the calculation to reduce it to the form like this [itex]I_0sin(\omega t + \theta)[/itex].
Doesn't [itex]R=X_L, X_C[/itex], mean that [itex]R=X_L=X_C\ ?[/itex]
 
  • #5
masterjoda said:
No just [itex]R=X_L[/itex].

Ah, my bad. I read it as both being R, rather than as part of a list of known values. Where's a semicolon when you need one? :smile:

Anyways, you can write the complex impedances directly:

ZR = R

ZL = 0 + jR

ZC = 0 - jXC

These can then be used in formulas in the usual way to find voltages and currents (complex). Convert the values to phasor form at the end.
 
  • #6
gneill said:
These can then be used in formulas in the usual way to find voltages and currents (complex). Convert the values to phasor form at the end.

I have used them and I got that last equation for [itex]I[/itex].
 
  • #7
I'm not sure why you've gone to the "raw" L and C versions of the impedances when you've been given XL and XC. You should be able to find the total current as a complex function of u, R, and XC.
 

FAQ: Solving circuit using complex numbers

1. What are complex numbers and how are they used in circuit analysis?

Complex numbers are numbers that consist of a real part and an imaginary part. In circuit analysis, complex numbers are used to represent the amplitude and phase of voltage and current signals in AC circuits.

2. How do you convert a circuit with resistors, capacitors, and inductors into a complex impedance circuit?

To convert a circuit with resistors, capacitors, and inductors into a complex impedance circuit, you can use the following formulas:

  • For resistors: ZR = R
  • For capacitors: ZC = 1/(jωC)
  • For inductors: ZL = jωL
Once you have converted all the components into complex impedance, you can use standard circuit analysis techniques to solve the circuit.

3. What is the phasor representation of a complex number and how is it used in circuit analysis?

The phasor representation of a complex number is a vector that represents the magnitude and phase of the complex number. In circuit analysis, phasors are used to analyze the behavior of AC circuits by converting time-domain signals into the frequency domain. This allows for easier calculations and analysis of circuit components.

4. How do you solve a circuit using complex numbers?

To solve a circuit using complex numbers, you can follow these steps:

  1. Convert all components (resistors, capacitors, and inductors) into their corresponding complex impedances.
  2. Apply Kirchhoff's laws to write out equations for the circuit.
  3. Use Ohm's law (V=IZ) and the complex impedance formulas to solve for the unknown variables.
  4. Convert the complex solutions back into their polar or rectangular form to get the final answer.

5. What are the advantages of using complex numbers in circuit analysis?

There are several advantages of using complex numbers in circuit analysis, including:

  • They allow for a more concise and efficient representation of signals in AC circuits.
  • They enable the use of phasors, which simplify calculations in the frequency domain.
  • They are necessary for analyzing circuits with reactive components (capacitors and inductors).
  • They can be used to analyze circuits with multiple frequencies.

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