What does this circuit diagram mean?

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Discussion Overview

The discussion revolves around understanding a circuit diagram that includes components such as capacitors and inductors, specifically focusing on their behavior as high pass filters. Participants express confusion about the nature of the diagrams and seek clarification on how to interpret them in the context of signal processing.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the structure of electrical diagrams, stating they expect them to be circular and closed.
  • Another participant explains that the diagrams represent parts of a circuit with a signal generator and an oscilloscope, suggesting that they can be analyzed as voltage dividers.
  • There is a discussion about the reactance of inductors and capacitors, with formulas provided for calculating their reactance based on frequency.
  • One participant notes that the output voltage behavior in the circuit differs from that of resistors in series, particularly in relation to the inductor's reactance increasing with frequency.
  • A participant seeks further clarification on how the circuit is completed and how to interpret the diagram, indicating a need for visual aids.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the circuit diagram, as some express confusion while others provide explanations. The discussion remains unresolved regarding the complete understanding of the circuit's function.

Contextual Notes

Participants mention limitations in understanding due to the nature of the diagrams and the need for visual representation to clarify concepts. There are also unresolved questions about the specific behavior of the circuit components as frequency changes.

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The attached picture shows four diagrams that I do not understand. To me electrical diagrams have to be circular and closed in nature, that's why these confuse me.

Once that question is answered I need to understand why II and III act as a high pass filter.

thanks
 

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The attached picture does not seem to be working. Could you try attaching it again?

Edit...The diagram has to be renamed as a PNG file. Now it works.

These drawings are just part of a circuit and assume you have something like a signal generator on the left and an oscilloscope on the right.

To work out what would happen in each case you regard the two devices as a voltage divider and work out what would happen to the reactance of the capacitor and inductor as you change frequency.

For example, the formula for reactance of an inductor is
XL = 2 * Pi * F * L where F is frequency and L is the inductance in Henries.
Can you see that if the inductance stayed the same, the reactance would go up as frequency goes up?
So, what would happen if the inductor was in a voltage divider? Depends if it was at the top or the bottom, doesn't it?

Then do the same for the capacitor
XC = reciprocal of ( 2 * Pi * F * C)

Note that the actual voltage you would get is not exactly like you would get with resistors in series. In diagram 1, the inductor has a higher reactance as the frequency increases so the output decreases.
Look up LCR networks somewhere and get more detail if you like.
 

Attachments

  • image-2.jpg
    image-2.jpg
    14.2 KB · Views: 730
Last edited:
vk6kro said:
The attached picture does not seem to be working. Could you try attaching it again?

Edit...The diagram has to be renamed as a PNG file. Now it works.

These drawings are just part of a circuit and assume you have something like a signal generator on the left and an oscilloscope on the right.

To work out what would happen in each case you regard the two devices as a voltage divider and work out what would happen to the reactance of the capacitor and inductor as you change frequency.

For example, the formula for reactance of an inductor is
XL = 2 * Pi * F * L where F is frequency and L is the inductance in Henries.
Can you see that if the inductance stayed the same, the reactance would go up as frequency goes up?
So, what would happen if the inductor was in a voltage divider? Depends if it was at the top or the bottom, doesn't it?

Then do the same for the capacitor
XC = reciprocal of ( 2 * Pi * F * C)

Note that the actual voltage you would get is not exactly like you would get with resistors in series. In diagram 1, the inductor has a higher reactance as the frequency increases so the output decreases.
Look up LCR networks somewhere and get more detail if you like.

sorry for the image... and thanks to vk6kro for fixing my mistake.

I don't understand how the circuit gets completed... does it look something like the image i have attached now (second image.jpg)[this one works, check on my windows box]

I understand the inductance I just don't understand how to interpret the diagram.

thanks Dominik
 

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  • second image.jpg
    second image.jpg
    16.4 KB · Views: 589
Yes, that is all.

It is just a part of a circuit shown like this to show the principle.
 
thanks
 

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