- #1
mathman44
- 204
- 0
Homework Statement
I can't seem to get a start on this. Could anyone provide a hint or something to get me started? Thanks...
Voltage divider; missing capacitor
- Thread starter mathman44
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SUMMARY
The discussion centers on the analysis of a voltage divider circuit involving complex impedances and sinusoidal inputs. Participants clarify that the output being a reduced copy of the input implies no phase difference for sinusoidal signals across all frequencies. The correct approach involves setting the "right" impedance to 1/9th of the "left" impedance, while also considering the conductivity, which is defined as the inverse of impedance. The specific conductivity values discussed include a "right" conductivity of 10-6 + jω10-10.
PREREQUISITES- Understanding of voltage divider circuits
- Familiarity with complex impedance and conductivity
- Knowledge of sinusoidal waveforms and their properties
- Basic principles of AC circuit analysis
- Study the principles of complex impedance in AC circuits
- Learn about the relationship between impedance and conductivity
- Explore the concept of phase difference in sinusoidal signals
- Investigate advanced voltage divider configurations and their applications
Electrical engineering students, circuit designers, and anyone involved in AC circuit analysis and voltage divider applications.
I can't seem to get a start on this. Could anyone provide a hint or something to get me started? Thanks...
- #2
mmmboh
- 401
- 0
I don't entirely get this questions either...does the fact that the output is an exact reduced copy mean that if we restrict ourselves to sinusoidal input, for any input frequency, there is no phase difference between the input and output waveforms?
- #3
mathman44
- 204
- 0
Anybody? I tried setting the "right" impedance equal to 1/9th of the "left" impedance to reform the voltage divider but this is a huge mess.
- #4
willem2
- 2,134
- 395
mathman44 said:
You can do it that way. It's easier to work with conductivity: 1/impedance
The "right" conductivity is 10^{-6} + j \omega 10^{-10}
The conductivity between A and B should be 9 times that.
- #5
Phrak
- 4,266
- 7
You want the imaginary parts of the complex impedences to be in the same ratio as the real parts; 9 to 1.
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