Passive Sign Convention in circuit

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Discussion Overview

The discussion revolves around the application of the passive sign convention in circuit analysis, specifically regarding the calculation of power in a voltage source within a given circuit. Participants explore the implications of voltage polarity, current direction, and the definitions of power generation and absorption in both voltage sources and resistors.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest using the formula P = V^2/R to calculate power, but express confusion over the implications of a positive power result for the voltage source, questioning whether it indicates the source is absorbing power.
  • Others clarify that the passive sign convention indicates that if current enters through the negative terminal of the voltage source, it generates power, while if it enters through the positive terminal, it dissipates power.
  • A participant emphasizes that power is defined as voltage times current, and the sign of power depends on the signs of voltage and current, with positive power indicating dissipation and negative power indicating delivery to the circuit.
  • Some participants note that an independent voltage source maintains a constant voltage across its terminals and that current must enter the resistor through its positive terminal, which influences the power calculations.
  • One participant mentions that independent sources can sometimes dissipate power in more complex circuits, suggesting the need for different analysis techniques in such cases.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of power signs in relation to the voltage source and the passive sign convention. There is no consensus on whether the voltage source is generating or absorbing power, and the discussion remains unresolved regarding the implications of the current's direction.

Contextual Notes

Some limitations in the discussion include the dependence on assumptions about current direction and voltage polarity, as well as the potential for independent sources to behave differently in complex circuits. These factors contribute to the uncertainty in the power calculations presented.

Air
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Homework Statement


Calculate the power in the voltage source for this given circuit:
CircuitDiagram.jpg



Homework Equations


P = \frac{V^2}{R} Or P = - \frac{V^2}{R}


The Attempt at a Solution


If we take the bottom left corner as the Initial node then and go around the circuit, then the voltage is a drop thus I thought we use: P = \frac{V^2}{R} = 10W however this means power is positive hence the voltage source is also an absorber which wouldn't make sense because there would be no generator in the circuit.
 
Last edited:
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Think about what the voltage source's polarity implies about the direction of the current.

Also, P = VI and Pabsorbed = Pgenerated
 
Air said:

Homework Statement


Calculate the power in the voltage source for this given circuit:
CircuitDiagram.jpg



Homework Equations


P = \frac{V^2}{R} Or P = - \frac{V^2}{R}


The Attempt at a Solution


If we take the bottom left corner as the Initial node then and go around the circuit, then the voltage is a drop thus I thought we use: P = \frac{V^2}{R} = 10W however this means power is positive hence the voltage source is also an absorber which wouldn't make sense because there would be no generator in the circuit.


The passive sign convention means that when current is entering the element through the negative end of the Voltage then the element is generating power and otherwise it is dissipating power.

So in your problem the independent Voltage source is generating power (as shown in the figure I have uploaded) and the resister is dissipating that power.

So for Voltage source

P (generated) = \frac{V^2}{R} =10W

and for Resistor

P(absorbed) = \frac{V^2}{R} =10W

I hope your problem is solved.
 

Attachments

  • Passive sign convention.jpg
    Passive sign convention.jpg
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Last edited:
Mastermind_14 said:
The passive sign convention means that when current is entering the element through the negative end of the Voltage then the element is generating power and otherwise it is dissipating power.

So in your problem the independent Voltage source is generating power (as shown in the figure I have uploaded) and the resister is dissipating that power.

So for Voltage source

P (generated) = \frac{V^2}{R} =10W

and for Resistor

P(absorbed) = \frac{V^2}{R} =10W

I hope your problem is solved.

It isn't given so how did you decide current is going from negative to positive in the voltage source? Also, should P=-10W as the voltage source is a generator?
 
Power is voltage times current, no matter what your reference node is. If voltage and current have the same sign, power is positive, else it is negative. Positive power means dissipated power and th is always the case in a passive resistor. Negative power is power delivered to the rest of the circuit. This can only happen with an active element, such as a power source.
 
I would like to add to what CEL has just posted that an Independent Voltage source always maintains the same voltage (In this case 10V) across its terminals with the same polarity. So this is how we get the signs of voltage across the Resistor (10 ohms).

The thing that we do now is to use the knowledge that a resistor is a passive element (always) so current must enter it through the +ve terminal of the voltage.

We also know that acording to KCL all the current through the resistor 'I' also passes through the Voltage Source (entering it from -ve side). Thus it is generating power.

Ho

NOTE: It is not necessary that an Independent source is always generating Power. In complicated circuits an independent source may even dissipate power. In that case you must analyze the circuit using different circuit analysis techniques. I will try to upload a circuit with a power dissipating Independent source here.
 
This is a circuit in which the Independent Voltage source and the resister are dissipating power while the Independent current source is generating Power.
 

Attachments

  • Two dissipating elements .jpg
    Two dissipating elements .jpg
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