Electrical Engineering: Circuit Problem

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Discussion Overview

The discussion revolves around a circuit problem involving the application of Ohm's law, Kirchhoff's Voltage Law (KVL), and Kirchhoff's Current Law (KCL) to determine the voltage Vx across a specific resistor in the circuit. Participants are focused on understanding the polarity of the leftmost 5-ohm resistor and its implications for current flow and voltage drops.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion regarding the polarity of the leftmost 5-ohm resistor, stating that the solution indicates the positive terminal is on the left and the negative on the right, which they believe is incorrect.
  • Another participant explains that when current enters a resistor, it loses voltage, and the positive sign indicates that the current had higher potential before entering the resistor, suggesting that an inverted polarity would imply the resistor is producing energy, which they argue is impossible.
  • A third participant notes that resistors produce thermal noise energy, which may imply a different understanding of energy interactions in resistors.
  • A later reply reiterates the initial confusion about the polarity and provides calculations for current and voltage drops across various resistors, ultimately concluding that the left side of the 5-ohm resistor should be at a higher potential than the right side.

Areas of Agreement / Disagreement

Participants do not reach a consensus regarding the polarity of the leftmost 5-ohm resistor, with some supporting the solution's indication of polarity while others express doubt. The discussion remains unresolved with competing views on the interpretation of voltage and current flow.

Contextual Notes

There are unresolved assumptions regarding the definitions of voltage polarity and energy absorption versus production in resistors. The calculations presented by participants depend on the interpretation of current flow and voltage drops, which may vary based on different circuit configurations.

softstyll
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Homework Statement


Consider the circuit shown in Figure Pl.64. Use repeated applications of Ohm's law, KVL, and KCL to eventually find Vx.


The Attempt at a Solution


I know how to solve the problem (almost). I'm having difficulty understanding the polarity at the left most resistor (5 ohms). I looked at the solution and it says that the positive terminal is on the left and the negative of the right of the resistor. I thought it was the opposite. The diagram of the circuit is attached.
 

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One way to look at it:

When current enters a resistor, it loses some voltage (potential). The + sign on the left of the 5ohm resistor simply means that the current had more potential before entering it than when it exits the resistor. If the polarity was inverted, it would mean the resistor would be producing energy instead of absorbing it, which is impossible (as far as I know).
 
Resistors do produce thermal noise energy.
 
softstyll said:

Homework Statement


Consider the circuit shown in Figure Pl.64. Use repeated applications of Ohm's law, KVL, and KCL to eventually find Vx.


The Attempt at a Solution


I know how to solve the problem (almost). I'm having difficulty understanding the polarity at the left most resistor (5 ohms). I looked at the solution and it says that the positive terminal is on the left and the negative of the right of the resistor. I thought it was the opposite. The diagram of the circuit is attached.

Why did you think that?
1A through 5Ohm = 5V across right most resistor
Hence 5V across the middle 5 Ohm resistor, so current through middle one = 1A
Similarly, 5V across 10Ohm resistor so current through it = 0.5Ohm
Thus total current through leftmost 5Ohm resistor = 2.5A
Voltage drop across 5Ohm = 5*2.5 = 12.5V
Thus Vx = 17.5V
Note that the current flows through leftmost 5V from left to right, Current always flows from higher potential to lower potential (opposite of direction of electron flow which is -ve to +ve) so left side of the 5Ohm resistor should be at higher potential than right side, or, the positive side is on the left and negative on the right
 

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