Finding Thevenin Equivalent Resistance: Circuit Analysis and Solution

AI Thread Summary
The discussion focuses on finding the Thevenin equivalent resistance in a circuit analysis problem. Participants clarify that the equivalent resistance on the right side simplifies to 12 kΩ, and the voltage across it can be calculated as 24 V from a 2 mA current. There is confusion regarding the polarity of the voltage across the resistors, with emphasis on conventional current flow from higher to lower potential. The open circuit voltage measured in the simulation is reported as 36 V, leading to further inquiries about the application of Kirchhoff's Voltage Law (KVL) in the analysis. The conversation highlights the importance of accurately determining voltage polarities and circuit configurations for correct calculations.
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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?
 

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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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