Method for always getting the sign of the power right

Question

In a circuit setup like the one below, I'm asked to find the power at the voltage source. I did:

$$KCL: I_V + I = I_R \iff I = I_R-I_V$$ $$I_R = \frac{V}{R} = \frac{1}{200} = 0.005 A$$ $$P_V = VI_V = V(I_R-I) = 10(0.005-0.003) = 20mW$$

But the answer is -20mW. I assumed the current to be flowing from the positive end of the voltage source to the node w/ resistance R and the current source I; and the current to be flowing from that node, through the Resistance (up to down vertically).

I know the energy given to the circuit has a negative sign by passive convention, but let's assume it was hard to deduce it from the circuit. From the analytical perspective (from these equations) how could I get the -20mW with a methodic approach? The only way out I can see is if the tension if flowing "out" from the "positive" end of the voltage source, it must have a negative sign (besides it having the same direction as the current I assumed).

What am I thinking wrong here and how can I do this without every be confused?

Answer 1

To prevent ohm's law being violated: -

I'm asked to find the power at the voltage source

Well, that's an ambiguous question that could mean "the power delivered by the voltage source" (in which case the answer is positive).

_{My advice in all of these types of question is not to jump to maths when common-sense is more useful.}

Answer 2

Forgive me for the length of this answer, I got carried away, but I hope this can be useful for beginners in general. The TLDR is at the bottom.

1. If current flows through some element in a direction from high potential end to low, then it is gaining (absorbing) energy. Otherwise it's losing (donating) energy.

2. Since resistors can never be energy sources, current must always flow into the terminal with the higher potential, and out of the lower potential end, in accordance with Ohm's law, and your current/polarity labels must be consistent with this principle.

3. For voltage and current sources, they may be either energy sinks or sources, and to determine which, you refer to point 1) above.

The trick to successful application of KVL, KCL and Ohm's law, is good labelling of the schematic, with current directions and voltage polarities. For resistors, you have no choice but to label in accordance with Ohm's law, but for voltage and current sources, it may not always be so simple. Take this small snippet for example:

^{simulate this circuit – Schematic created using CircuitLab}

I take care labelling resistor R5, current direction through it must be consistent with point 2 above, with higher potential being where current enters, from the right, since resistors always absorb energy.

Voltage source V1 deserves your attention. I've included a small current direction and value above it to illustrate the usefulness of point 1 above. The product of \$I_3\$ and \$V_1\$ tells you the power being absorbed by that source, just as the product of \$V_5\$ and \$I_3\$ is the power being absorbed by R5. The important thing about current in V1 is that it is flowing from higher potential (right) end to lower (left), just like a resistor would do, and just like a resistor, V1 is absorbing energy.

Now turn your attention to V2. Current is still flowing right to left, but now hypothetical positive charges (remember conventional current) are gaining 5eV of potential energy on their journey through the source, towards the higher potential. In other words, that source V2 is charging up, receiving energy, instead of being drained.

Now I want to point out that I have added a small label for current above V2, and this is to help me work out the power later. Notice how its value is \$I_3\$ negated, and flowing left to right. It means the same thing, of course, as positive \$I_3\$ flowing right to left, but this extra label is consistent with point 2 above, which means when I multiply the two values \$-I_3\$ and \$V_2\$, I will have the power being absorbed. If that absorption is positive, energy is being received. If it's negative, V2 is donating energy to the circuit.

Now tackle your problem with all labels in place. Let's pretend I got really lucky, and happened to guess polarities and current directions correctly:

^{simulate this circuit}

Here I've "guessed" \$I_2\$ to be flowing clockwise in the left loop, but I notice that for component V2 this is contrary to the principle of current flowing from high to low potential. So I added a little reminder, by negating the current's direction and magnitude next to its symbol, which doesn't in any way violate the "truth" of that current. Now, whatever value we get for \$I_2\$, we know that the power being dissipated in V1 will be \$P_{V2} = -I_2 \times V_2\$.

For completeness, I'll analyse the circuit, and apply this little trick in the calculation of \$P_{V2}\$. Start with KCL at node A:

$$ I_2 + I_1 = I_3 $$

Now KVL for the left loop. Starting at node A, and going clockwise, I encounter a decrease in potential (according to my labels, always, always obey the labels) as I traverse R1, and then a 10V increase as I jump across V2, arriving back where I started:

$$ -(I_3R_3) + 10 = 0 $$

For the right loop, going clockwise, we drop in potential while crossing I1, and increase as we traverse R1:

$$ -V_1 + (I_3R_3) = 0 $$

We really didn't need to do a complete nodal analysis, because this problem is trivial, but solving these of course yields:

$$ \begin{aligned} I_2 &= 2mA \\ \\ I_3 &= 5mA \\ \\ \end{aligned} $$

And now for the answer to your question, the power being dissipated in V2. Well, thanks to our current label next to V2, which conforms to our requirement that for power calculations, current always be considered flowing from high potential to low:

$$ \begin{aligned} P_{V2} &= -I_2 \times V_2\\ \\ &= -2mA \times +10V \\ \\ &= -20mW \end{aligned} $$

The fact that we have a negative value for power indicates that voltage source V2 is losing energy, donating it to the rest of the circuit.

Finally, notice how the labels for current direction and voltage across current source I1 are not in accordance with point 2. Therefore the product of 3mA and V1 is not a good indicator of power dissipation in I1. The correct calculation would be:

$$ \begin{aligned} P_{I1} &= -I_1 \times V_1\\ \\ &= -3mA \times +10V \\ \\ &= -30mW \end{aligned} $$

This negative result tells you that the current source is also an energy source.

TLDR

Power calculations are always performed according to the convention that current flows within an element from its higher potential terminal to lower. If this convention is followed, then positive values for power indicate the element is an energy sink, absorbing energy. Negative values indicate an energy source, losing energy, donating it to the rest of the circuit.

If we label the circuit as follows (note the direction of \$I_2\$):

^{simulate this circuit}

then a nodal analysis would reveal \$I_2 = -2mA\$. The resulting power calculation is:

$$ \begin{aligned} P_{V2} &= I_2 \times V_2 \\ \\ &= -20mA \times +10V \\ \\ &= -20mW \end{aligned} $$

Because our chosen current direction is in accordance with the princple that current flows from high to low potential through V2, we may trust the sign of the product of current through and voltage across V2 to correctly indicate the power dissipation in V2. This negative value tells us that voltage source V2 is an energy source, donating to the rest of the circuit.