Finding Thevenin Equivalent Resistance: Circuit Analysis and Solution

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent resistance in a given circuit, focusing on circuit analysis, voltage calculations, and the implications of polarity in resistor voltage. Participants explore various approaches to determine the open circuit voltage and equivalent resistance, while addressing potential misunderstandings in the application of circuit theory.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the equivalent resistance of the right side of the circuit as 12 kΩ and finds a voltage of 24 V from a current of 2 mA.
  • Another participant questions the polarity of the voltage, suggesting the bottom wire should be considered at 0V.
  • Further clarification is sought regarding the polarity of the voltage across the 12 kΩ resistor, particularly in relation to the direction of current flow.
  • Some participants discuss the implications of assuming a polarity for the voltage, noting that it can be corrected later if necessary.
  • There is a mention of the open circuit voltage being 36V, with one participant expressing confusion about the 12V source's role in the circuit.
  • Participants reference the use of Kirchhoff's Voltage Law (KVL) to analyze the circuit, leading to differing interpretations of the voltage across components.

Areas of Agreement / Disagreement

Participants express differing views on the polarity of voltages and the implications of current direction, indicating a lack of consensus on these aspects. The discussion remains unresolved regarding the correct interpretation of the circuit elements and their interactions.

Contextual Notes

Some assumptions about the circuit configuration and the role of the voltage sources are not fully clarified, leading to potential misunderstandings in the calculations presented. The discussion also reflects varying interpretations of conventional current flow and its impact on voltage polarity.

garr6120
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Homework Statement


given the circuit shown below find the the vanin equivalent at resistance R.

Screen Shot 2017-10-07 at 11.34.55 PM.png

Homework Equations


the resistances on the right side of the circuit can be simplified: 18 || (24+12) = 12 kΩ.

The voltage can be found from 2 mA * 12 kΩ = 24 V.

The voltage Va = 12-24 = -12 V

The Attempt at a Solution



When using multisim and the multimeter within the program it tells me the voltage of the open circuit is 36V.

Can anyone help me out?
 

Attachments

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garr6120 said:
The voltage can be found from 2 mA * 12 kΩ = 24 V.
With what polarity? Assume the bottom wire to be at 0V.
 
cnh1995 said:
With what polarity? Assume the bottom wire to be at 0V.

- V + with the - charge facing the 0V wire
 
garr6120 said:
- V + with the - charge facing the 0V wire
What is V?

What is the polarity of the bottom of the 12k resistance if the current through it goes from bottom to top?
 
cnh1995 said:
What is V?

What is the polarity of the bottom of the 12k resistance if the current through it goes from bottom to top?

+ to - from left to right.
 
garr6120 said:
+ to - from left to right.
I was talking about the equivalent resistance of the three resistors in the rightmost loop, which is 12k ohm. Draw this simplified circuit and find the polarity of the resistor terminal connected to the 0V point.
 
cnh1995 said:
I was talking about the equivalent resistance of the three resistors in the rightmost loop, which is 12k ohm. Draw this simplified circuit and find the polarity of the resistor terminal connected to the 0V point.

Screen Shot 2017-10-08 at 2.06.49 PM.png
 
garr6120 said:
The current source is of 2mA, and you need to find the open circuit voltage. You have instead short-circuited the resistance R.
 
cnh1995 said:
The current source is of 2mA, and you need to find the open circuit voltage. You have instead short-circuited the resistance R.

Screen Shot 2017-10-08 at 2.53.25 PM.png
 
  • #10
garr6120 said:
The polarity of the resistor voltage is wrong. Conventional current flows from higher potential to lower potential.
 
  • #11
cnh1995 said:
The polarity of the resistor voltage is wrong. Conventional current flows from higher potential to lower potential.

Even if it is i can correct it when I am done because its an arbitrary polarity. If i get negative for my answer I have to flip the polarity at the end so that if i recalculate my final answer the answer will be positive. So its just a prior guess right now.

but here it is.
Screen Shot 2017-10-08 at 3.06.50 PM.png
 
  • #12
garr6120 said:
but here it is.
195683-90ad819e6d3254762272b98b53f12a3e.png
That is correct. So now you can see why the answer is 36V.
 
  • #13
cnh1995 said:
That is correct. So now you can see why the answer is 36V.

I thought the 12 V source wouldn't receive any current because of the open circuit.

Should I use KVL because when i make a loop at the open circuit i get:
-24V-2mA*12Ω=-12V
 

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