Recognitions:
Gold Member

Passive Sign Convention

1. The problem statement, all variables and given/known data
Calculate the power in the voltage source for this given circuit:

2. Relevant equations
$$P = \frac{V^2}{R}$$ Or $$P = - \frac{V^2}{R}$$

3. 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 sence because there would be no generator in the circuit.
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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

 Quote by Air 1. The problem statement, all variables and given/known data Calculate the power in the voltage source for this given circuit: 2. Relevant equations $$P = \frac{V^2}{R}$$ Or $$P = - \frac{V^2}{R}$$ 3. 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 sence because there would be no generator in the circuit. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution

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.
Attached Thumbnails

Recognitions:
Gold Member

Passive Sign Convention

 Quote by Mastermind_14 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. Attached Thumbnails