Find Thevenin Equivalent Circuit - 571.43 Ohms & 1000V

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent circuit for a given electrical circuit, focusing on determining both the Thevenin resistance and voltage. Participants explore various methods and calculations related to the circuit's components, including resistors in series and parallel configurations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the Thevenin resistance as 571.43 ohms, derived from combining resistors in parallel and series.
  • Another participant questions the validity of adding currents together in parallel, emphasizing the need to confirm that no series resistances interfere.
  • Multiple participants calculate the current through the circuit, with one stating it as 0.005 A and another later correcting it to a smaller value in milliamps.
  • Concerns are raised about the magnitude of the calculated voltage, with one participant expressing confusion over obtaining a high voltage value and questioning the correctness of their calculations.
  • Corrections are made regarding earlier miscalculations, with participants updating their figures as they clarify their reasoning.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations, as there are multiple interpretations and corrections of the steps involved. Disagreements persist regarding the proper approach to adding currents and the resulting voltage calculations.

Contextual Notes

Some calculations depend on the correct interpretation of series and parallel configurations, and there are unresolved steps in the reasoning that may affect the final outcomes.

illidari
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Homework Statement



thevinen.jpg


Homework Equations



Find equivalent thevenin circuit

The Attempt at a Solution



I found the equivalent thevenin resistor already.
The 1000 + 500 are in parallel. Come out to 333.333 equivalent.
Equivalent is in series with 1000, then its 1333.333
1333.33+1000 in parallel is 571.43 ohms

Thevenin resistor equivalent = 571.43 ohms

I am stuck on the thevenin voltage.

From my understanding I can take the 5V and divide it by the 1k resistor to get the current at that point. Which is .005A

Am I allowed to add the currents together since they are in parallel? Make it 2.005 A?

Reduce the resistors down to one: 571.43 ohms

V=IR ...571.43 ohms * 2.005 amp = 1000+ Volts...this seems horribly wrong.

Help :(?
 
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You have to make sure that currents really are in parallel before you can add them; No series resistances in the way!

One way to approach this problem would be to do multiple transformations (Thevenin, Norton, as required), working your way from left to right through the circuit. You already began that process when you proposed converting the 5V supply and its series 1k resistance to a current supply with a parallel 1k resistance.
 
5V / 1k = .005 A

1k + 500 resistor in parallel = 333.33 ohms

333.33 ohms * .005 A = 1.666V

333.33ohms + 1k ohms in series = 1333.33 ohms

1.666V/1333.33ohms = 1.25 mA

Nothing in series between the currents so current is 1.25mA+2A = 2.00125A

1.33k + 1k in parallel = 571.43 ohms

571.43 ohms * 2.00125A = 1143.57428 V

Should I be expecting something this massive in voltage? Feels like I am taking the wrong second step once I get the .005A
 
Last edited:
illidari said:
5V / 1k = .005 A

1k + 500 resistor in parallel = 333.33 ohms

333.33 ohms * .005 A = 1.666V

333.33ohms + 1k ohms in series = 1333.33 ohms

1666.66V/1333.33ohms = 1.25 A
How did you get from 1.666V to 1666.66V ?
 
Sorry misread my paper when I was working it out.

Was suppose to still be 1.667 which reduces the the current I got to mA, not A. I updated above.
 
Okay, looks better now. The voltage is okay for that amount of current being pushed through several hundred Ohms :smile:
 

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