# Two seemingly analogous cases of using Kirchhoff's current law to calculate current through a node

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. 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. 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).

• 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 $9\:\text{A}$".
– jonk
Commented Mar 22, 2022 at 0:20
• Look over your own work, compare it with mine, and tell me where you went wrong.
– jonk
Commented Mar 22, 2022 at 0:27
• 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.
– jonk
Commented Mar 22, 2022 at 1:04
• 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.
– jonk
Commented Mar 22, 2022 at 1:04

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
• Where is the assumption $i_o = v_o/2$ from? Commented Mar 22, 2022 at 0:38
• @jonk Oh, so they're using the $2 \ \Omega$ resistor to find $i_o$ using Ohm's law? Commented Mar 22, 2022 at 0:48