# Passive Sign Convention in circuit

• Engineering
• Air
In summary, the power in the voltage source for this given circuit is calculated using the equations P = \frac{V^2}{R} or P = - \frac{V^2}{R}. By using the passive sign convention, it is determined that the voltage source is generating power with a value of 10W, while the resistor is dissipating the same amount of power. The current direction is determined based on the polarity of the voltage source and the fact that a resistor is a passive element. It should be noted that in more complex circuits, the power generated or absorbed by an independent source may vary.
Air

## Homework Statement

Calculate the power in the voltage source for this given circuit:

## 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:
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:

## 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
13.2 KB · Views: 1,027
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
11 KB · Views: 788

## 1. What is the Passive Sign Convention in circuit?

The Passive Sign Convention is a set of rules used to describe the direction of current and voltage in a circuit. It states that the current entering a passive circuit element (such as a resistor, capacitor, or inductor) is always considered to be positive, and the voltage drop across the element is always considered to be negative.

## 2. Why is the Passive Sign Convention important in circuit analysis?

The Passive Sign Convention allows for consistent and standard descriptions of circuits, making it easier to analyze and understand complex circuits. It also helps to avoid errors and confusion when solving circuit equations.

## 3. Can the Passive Sign Convention be applied to all types of circuits?

Yes, the Passive Sign Convention can be applied to all types of circuits, including DC and AC circuits. It is a fundamental principle in circuit analysis that is used to simplify and solve circuit equations.

## 4. What happens if the Passive Sign Convention is not followed in circuit analysis?

If the Passive Sign Convention is not followed, it can lead to incorrect results and confusion when solving circuit equations. It is important to adhere to this convention to maintain consistency and accuracy in circuit analysis.

## 5. Are there any exceptions to the Passive Sign Convention?

Yes, there are some exceptions to the Passive Sign Convention, such as with active circuit elements like transistors and diodes. In these cases, the direction of current and voltage can vary depending on the specific circuit configuration.

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