Help on thevenizing a Scheering Bridge

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In summary, the conversation discusses the process of "thevenizing" a Scheering Bridge under specific conditions and the steps involved in determining the magnitude and phase of the sinusoidal current flowing through a galvanometer with negligibly small internal resistance. The method presented involves calculating the Thevenin equivalent circuit and using an expression involving jw to determine the phase, which is not always 90 degrees and depends on the frequency.
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Help on "thevenizing" a Scheering Bridge

Homework Statement



Under the following conditions: C1=C, C2=0, Cx=2C, R2=R3=R, Rx=0, find the magnitude and phase of the sinusoidal current flowing through the galvanometer D (ID) whose internal resistance is neglibly small.

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The Attempt at a Solution

We have an unbalanced condition. We can determine ID by means of the Thevenin equivalent circuit:

Steps:
1) get Vth:

VTH = VCB = VC - VB = VAC - VAB

(Z1= 1/jwC, Zx=1/jw2C, Z3=Z2=R)

2) Now let's calculate Zth, which will have a real part (resistor) and imaginary part (capacitor)

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ZTH = (Z1 || Z2) + (Zx || Z3)

3) Finally let's form the Thevenin equivalent circuit:

2jescqg.png


ID = VTH / ZTH

ok, up to this point I think we have got the magnitude of ID. What about the phase ? Since a capacitor delays voltage 90 degrees, I believe that the phase of ID is 90, but I am not sure...

Would this work as a solution?

Thanks in advance...
 
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Your method will work. The expressions involved are rather complicated and you will have to be very careful with your algebra to get the final result, which is surprisingly simple.

The phase is not 90 degrees; it depends on the frequency. You will have to derive an expression involving jw: Arctan(Imaginarypart/Realpart)
 
  • #3


The Electrician said:
Your method will work. The expressions involved are rather complicated and you will have to be very careful with your algebra to get the final result, which is surprisingly simple.

The phase is not 90 degrees; it depends on the frequency. You will have to derive an expression involving jw: Arctan(Imaginarypart/Realpart)

Thanks for your comment! I also thought that the calculations were very complicated...
 

Related to Help on thevenizing a Scheering Bridge

1. What is Thevenizing a Scheering Bridge?

Thevenizing a Scheering Bridge is a process used in electrical circuit analysis to simplify and solve complex networks. It involves replacing a portion of the circuit with an equivalent Thevenin circuit, which is a simplified circuit with a single voltage source and a single resistor.

2. Why is Thevenizing a Scheering Bridge useful?

Thevenizing a Scheering Bridge is useful because it allows for a complex circuit to be simplified into a single equivalent circuit, making it easier to analyze and solve. It also allows for the determination of the voltage and current at any point in the circuit.

3. What is the procedure for Thevenizing a Scheering Bridge?

The procedure for Thevenizing a Scheering Bridge involves identifying the portion of the circuit to be replaced with an equivalent Thevenin circuit, finding the open circuit voltage and the equivalent resistance, and then constructing the Thevenin circuit by connecting the voltage source and resistor in series.

4. What are the advantages of using Thevenizing a Scheering Bridge?

Thevenizing a Scheering Bridge has several advantages, including simplifying complex circuits, reducing the number of components needed for analysis, and allowing for easy determination of voltage and current at any point in the circuit. It also helps in studying the behavior of the circuit under different conditions.

5. Are there any limitations to Thevenizing a Scheering Bridge?

While Thevenizing a Scheering Bridge is a useful tool, it does have certain limitations. The circuit must be linear and contain only resistors, voltage sources, and current sources. It also cannot be used for circuits with dependent sources or with time-varying signals.

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