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I'm currently studying the textbook Fundamentals of Electric Circuits, 7th edition, by Charles Alexander and Matthew Sadiku. Chapter 2.4 Kirchhoff's Laws has the following practice problems:

Find current \$i_o\$ and voltage \$v_o\$ in the circuit shown in Fig. 2.25. enter image description here Solution: Applying KCL to node \$a\$, we obtain $$3 + 0.5 i_o = i_o \ \Rightarrow i_o = 6 \ \text{A}$$ For the \$4 \ \Omega\$ resistor, Ohm's law gives $$v_o = 4 i_o = 24 \ \text{V}$$

Find \$v_o\$ and \$i_o\$ in the circuit of Fig. 2.26. enter image description here Answer: \$12 \ \text{V}, 6 \ \text{A}\$.

For Figure 2.26, it seems to me that \$i_o\$ is the current through the node above the \$2 \ \Omega\$ resistor. This node has a \$9 \ \text{A}\$ current flowing into it and a \$0.25i_o\$ current flowing out of it. Therefore, by Kirchhoff's current law, we have that \$9 \ \text{A} - 0.25i_o = i_o \ \Rightarrow i_o = 7.2 \ \text{A}\$. But this is incorrect. So what am I doing wrong here? This problem seems to be analogous to the previous one, so I don't understand what's different (besides the additional branch).

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  • \$\begingroup\$ The KCL is easy: \$\frac{V}{2}+\frac{V}{8}+0.25\cdot\frac{V}{2}=9\:\text{A}\$. This trivially solves out as \$V=12\:\text{V}\$. Ground the bottom wire. Then this is just: "The current out of the \$2\:\Omega\$ resistor, plus the current out of the \$8\:\Omega\$ resistor, plus the current through the dependent current source equals \$9\:\text{A}\$". \$\endgroup\$
    – jonk
    Mar 22, 2022 at 0:20
  • \$\begingroup\$ Look over your own work, compare it with mine, and tell me where you went wrong. \$\endgroup\$
    – jonk
    Mar 22, 2022 at 0:27
  • \$\begingroup\$ Let's assume that the lower node is called \$V_{_\text{A}}\$ and the upper node is called \$V_{_\text{B}}\$ and that the voltage difference between them is \$v_o=V_{_\text{B}}-V_{_\text{A}}\$ per their definition. \$\endgroup\$
    – jonk
    Mar 22, 2022 at 1:04
  • \$\begingroup\$ Then \$i_o=\frac{V_{_\text{B}}-V_{_\text{A}}}{2\:\Omega}\$ and these two KCL statements: \$\frac{V_{_\text{B}}}{2\:\Omega}+\frac{V_{_\text{B}}}{8\:\Omega}+0.25\cdot i_o=\frac{V_{_\text{A}}}{2\:\Omega}+\frac{V_{_\text{A}}}{8\:\Omega}+9\:\text{A}\$ and \$\frac{V_{_\text{A}}}{2\:\Omega}+\frac{V_{_\text{A}}}{8\:\Omega}+9\:\text{A}=\frac{V_{_\text{B}}}{2\:\Omega}+\frac{V_{_\text{B}}}{8\:\Omega}+0.25\cdot i_o\$. These solve out as \$i_o=6\:\text{A}\$ and \$V_{_\text{A}}=V_{_\text{B}}-12\:\text{V}\$. You can assign anything you want to either \$V_{_\text{A}}\$ or \$V_{_\text{B}}\$. But only one. \$\endgroup\$
    – jonk
    Mar 22, 2022 at 1:04

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The major issue in your attempted solution is that you are in no way taking into account the current through the 8 Ohm resistor. Remember that current into a node must equal the current out of a node. Summing currents at the top node, we have: $$9 = \frac{v_o}{2} + 0.25 \cdot \frac{v_o}{2} + \frac{v_o}{8}$$ where $$i_o = \frac{v_o}{2}$$ This quickly simplifies to: $$9 = \frac{3 \cdot v_o}{4}$$ At which point solving for current and voltage becomes trivial. Hopefully this clarifies things

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  • \$\begingroup\$ Where is the assumption \$i_o = v_o/2\$ from? \$\endgroup\$ Mar 22, 2022 at 0:38
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    \$\begingroup\$ @ThePointer It comes from Ohm's Law. Why do you think it is an assumption rather than an expression resulting from such a simple law? Don't you see that the same voltage across the 8 Ohm resistor must also be exactly the same across the 2 Ohm resistor, too? \$\endgroup\$
    – jonk
    Mar 22, 2022 at 0:47
  • \$\begingroup\$ @jonk Oh, so they're using the \$2 \ \Omega\$ resistor to find \$i_o\$ using Ohm's law? \$\endgroup\$ Mar 22, 2022 at 0:48
  • \$\begingroup\$ @ThePointer Yes!! In a sense, anyway. \$\endgroup\$
    – jonk
    Mar 22, 2022 at 0:48
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    \$\begingroup\$ @ThePointer No, because io doesn't flow through the 8 ohm resistor. What makes you think it does? Just think about it. If io flows through the 2 ohm resistor then the vo = 2 * io and this is also the voltage across the 8 ohm resistor. Hence the current through the 8 ohm resistor will be vo / 8 = io / 4 \$\endgroup\$ Mar 22, 2022 at 2:17

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