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

Question

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).

Answer 1

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