# Understanding the reasoning behind these power calculations involving voltage and current sources in circuits

I'm currently studying the textbook Fundamentals of Electric Circuits, 7th edition, by Charles Alexander and Matthew Sadiku. Chapter 1.6 Circuit Elements gives the following example and practice problem:

Example 1.7
Calculate the power supplied or absorbed by each element in Fig. 1.15. Solution:
...
For $$\p_4\$$, we should note that the voltage is 8 V (positive at the top), the same as the voltage for $$\p_3\$$ since both the passive element and the dependent source are connected to the same terminals. (Remember that voltage is always measured across an element in a circuit.) Since the current flows out of the positive terminal, $$p_4 = 8(-0.2I) = 8(-0.2 \times 5) = -8 \ \text{W} \ \ \ \text{Supplied power}$$

Practice Problem 1.7
Compute the power absorbed or supplied by each component of the circuit in Fig. 1.16. Answer: $$\p_1 = -225 \ \text{W}\$$, $$\p_2 = 90 \ \text{W}\$$, $$\p_3 = 60 \ \text{W}\$$, $$\p_4 = 75 \ \text{W}\$$.

I'm not totally clear on the reasoning behind the power calculations for element $$\p_4\$$ of example 1.7 and element $$\p_3\$$ of practice problem 1.7. My understanding is that, for element $$\p_4\$$ of example 1.7, we have that $$\0.2I = 0.2(-5 \ \text{A})\$$ because the current $$\I = 5 \ \text{A}\$$ is running counter-conventional, from the positive terminal of the $$\20 \ \text{V}\$$ ideal independent voltage source to its negative terminal. But why is the voltage over $$\p_4\$$ the $$\8 \ \text{V}\$$ over element $$\p_3\$$? Furthermore, for element $$\p_3\$$ of practice problem 1.7, we see that the current $$\I = 25 \ \text{A}\$$ is running in the conventional direction, from the negative terminal of the $$\2 \ \text{V}\$$ ideal independent voltage source to its positive terminal, so we have $$\0.12I = 0.12(25 \ \text{A})\$$. But why is the current through $$\p_3\$$ $$\20 \ \text{A}\$$, giving us a power of $$\0.12(25 \ \text{A}) \times 20 \ \text{A} = 60 \ \text{W}\$$, rather than the $$\I = 25 \ \text{A}\$$, which would then give us $$\0.12(25 \ \text{A}) \times 25 \ \text{A} = 75 \ \text{W}\$$?

To clarify, the textbook has not yet introduced the concept of a component being in parallel vs being in series, so the reader would not know this at this point in the textbook.

## 1 Answer

For $$\p_4 \$$ in example 1.7 the current flows from the lower electric potential to the higher electric potential. It flows through a voltage increase. So the power dissipated in $$\p_4 \$$ is $$P_{p4}=I_{p4}V_{p4} = (0.2\frac{\text{A}}{\text{A}} \cdot 5\text{A}) \cdot (-8\text{V}) = -8\text{W} \: \: \: \text{(power delivered)}$$

$$\p_3 \$$ and $$\p_4 \$$ have the same voltage drop across them, because they are in parallel connection.

For $$\p_3 \$$ in practice problem 1.7 the current flows from the higher electric potential to the lower electric potential. It flows through a voltage drop. So the power dissipated in $$\p_3 \$$ is $$P_{p3} = I_{p3}V_{p3}=(20\text{A}) \cdot (0.12 \frac{\text{V}}{\text{A}}\cdot25\text{A}) =20\text{A} \cdot 3\text{V}=60\text{W} \: \: \: \text{(Power absorbed)}$$

Where the current comes from earlier in the circuit has no influence on the power dissipated in a particular passive circuit component. The only thing that matters are the current direction, and the voltage polarity - and the magnitudes of course.

• Why do you have $\frac{\text{A}}{\text{A}}$ and $\frac{\text{V}}{\text{A}}$? So your justification for $p_3$ and $p_4$ having the same voltage drop across them is that they are in parallel connection? The confusing thing is that the authors have not yet introduced a distinction between parallel and series (the concept of components being in parallel vs being in series has not yet been introduced), so there's no way for me to know this at this point in the textbook. Nov 19 '21 at 15:34
• @ThePointer $p_4$ is a current-controlled current source. $p_3$ is a current-controlled voltage source. Think about what that means, then the units I included in the calculation make sense. With regards to your second point, you don't have to know about series or parallel connections to see that they have the same voltage drop. You can see it with a quick use of KVL (Kirchoff's voltage law) or just notice both components' terminals are connected to the two same circuit nodes.
– Carl
Nov 19 '21 at 15:42
• "notice both components' terminals are connected to the two same circuit nodes" Yes, I think that was the justification that the authors wanted the reader to use for that one. The image in this en.wikipedia.org/wiki/Node_(circuits) Wikipedia article illustrates this. Nov 19 '21 at 15:48
• @ThePointer Okay, thanks for letting me know.
– Carl
Nov 19 '21 at 15:58