Finding Thevenin Equivalent of Circuit

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent of a given electrical circuit. Participants analyze the circuit's resistances and voltages, focusing on the methods for calculating the Thevenin resistance (Rth) and the open circuit voltage.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the Thevenin resistance is 5 ohms, based on their analysis of the circuit with the current source set to zero.
  • Another participant challenges this analysis, stating that the rightmost resistor is in series with the parallel combination of two 2-ohm resistors, suggesting a different approach to calculating Rth.
  • A later reply indicates that the correct Rth should be calculated as 4//2 + 2, leading to a value of 3.33 ohms, while also noting that the rightmost resistor does not carry current due to the open circuit.
  • Participants discuss the open circuit voltage calculation using a current divider, with varying interpretations of how the current divides among the resistors.

Areas of Agreement / Disagreement

There is disagreement regarding the calculation of the Thevenin resistance and the open circuit voltage. Multiple competing views remain, and the discussion has not reached a consensus on the correct approach or values.

Contextual Notes

Participants express uncertainty about the correct configuration of resistors and the implications of the open circuit on current flow, which affects their calculations. There are unresolved mathematical steps in the voltage calculations and differing interpretations of the circuit layout.

johndoe
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3104980940_408866c01d_o.jpg


Finding the thevenin equivalent of the above circuit


This is what I did, I set the current source to zero, no current passes through
so the Rth is the resistor on farthest right // with another 2 in series with 4.
so it is 5ohms?

Finding the open circuit voltage, I use a current divider(the current source is 10A given):
10 times ( 1/(2//2+2)) ) / (1/2 + 1/(2//2+2)) ) does it look alright to you guys? others solutions are appreciated
 
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johndoe said:
3104980940_408866c01d_o.jpg


Finding the thevenin equivalent of the above circuit


This is what I did, I set the current source to zero, no current passes through
so the Rth is the resistor on farthest right // with another 2 in series with 4.
so it is 5ohms?

Finding the open circuit voltage, I use a current divider(the current source is 10A given):
10 times ( 1/(2//2+2)) ) / (1/2 + 1/(2//2+2)) ) does it look alright to you guys? others solutions are appreciated

Your analysis is wrong. With the source open circuited, you have the rightmost resistor in series with the parallel of 2 and 4 ohm.
For the open circuit voltage, the source divides by a 2 ohm resistor on the left and 2+2 = 4 ohm in the right. The voltage is the current in the right arm times 2 ohm.
 
CEL said:
Your analysis is wrong. With the source open circuited, you have the rightmost resistor in series with the parallel of 2 and 4 ohm.
For the open circuit voltage, the source divides by a 2 ohm resistor on the left and 2+2 = 4 ohm in the right. The voltage is the current in the right arm times 2 ohm.

o so the Rth is 4//2 +2= 3.33 and since there is no current in the far right resistor, the current divider just involves the other 3 ?
 
hello johndoe,
yes the current divides between the three resistor and does not involve the rightmost resistor as no current flows because of open circuit.
And your Rth looks correct to me
 

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