Finding Voltage at Point 2 Using Superposition Method

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Discussion Overview

The discussion revolves around finding the voltage at point 2 in a circuit using the superposition method. Participants analyze the circuit by considering scenarios with no voltage source and no current source, employing node voltage analysis and voltage divider principles.

Discussion Character

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

Main Points Raised

  • One participant presents their calculations for the voltage at point 2, obtaining -4 volts for the circuit without a voltage source and 2 volts for the circuit without a current source, leading to a combined result of -2 volts.
  • Another participant questions the validity of the voltage divider calculation, suggesting that the voltage across the resistors in series should be considered differently.
  • There is a discussion about the application of the voltage divider formula, with some participants asserting that the total voltage across series resistors is divided, while others suggest that the voltage divider works 'oppositely' to the current divider.
  • A participant offers moderate assistance, indicating that the original poster seems to understand the problem but may be overlooking certain details.
  • Clarification is provided regarding the voltage across the 20-ohm resistor in the circuit without the current source, which is acknowledged by the original poster.

Areas of Agreement / Disagreement

Participants express differing views on the application of the voltage divider formula and the interpretation of the circuit analysis. No consensus is reached regarding the correct approach to calculating the voltage at point 2.

Contextual Notes

Some assumptions regarding the circuit configuration and the application of circuit analysis techniques may be missing or unclear. The discussion reflects varying interpretations of the voltage divider and node voltage analysis methods.

Tekneek
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Homework Statement



I have to find voltage at point 2 using superposition method. So, I drew one circuit with no voltage source and another with no current source, and then tried to analyze each one separately.

(On the picture starting from left, its 10 volts, then 10ohms, then at top 5 ohms, then 1 amp current source pointing down, then on far right 20 ohms)

The Attempt at a Solution



For No Voltage Source:

Using Node Voltage analysis at V2:
(V2-0)/5 + 1 + (V2-0)/20 = 0
V2= -4 Volts

For No current Source:

Using Voltage divider formula:
V2 = (5/(5+20))*10 = 2 Volts

Adding the two V2's i get V2 = -2, which i think is wrong.
 

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Hi Tekneek. This line doesn't look right ...

V2 = (5/(5+20))*10
 
NascentOxygen said:
Hi Tekneek. This line doesn't look right ...

For the no current source circuit isn't V1 10 volts? Then since 5ohms and 20ohms were in series i used the voltage divider formula. Also the 10 ohms resistor and 5+20 ohms are in parallel so they would have the same voltage (10v)?
 
Voltage divider works 'oppositely' to current divider.
 
Last edited:
FOIWATER said:
Voltage divider works 'oppositely' to current divider

What do you mean by that? Isn't the total voltage across the resistors in series divided?
 
The rules require I can only offer moderate assistance. Based on where you managed to get in this problem, I think you understand this but are simply not seeing it. Look at what I have attached

**edit that 30 should be a 20, my eyes are not good
 

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FOIWATER said:
The rules require I can only offer moderate assistance. Based on where you managed to get in this problem, I think you understand this but are simply not seeing it. Look at what I have attached

**edit that 30 should be a 20, my eyes are not good

Oh ok so V2 is the voltage across 20ohms resistor (for circuit with no current source). Thank You!
 
Yeah that should make sense looking at where you have placed the ground, right?

You're welcome, I am glad you came to that conclusion on your own as well!
 

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