Use nodal analysis to find Io in the network in Fig. P3.11.

#### Solution:

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We will “bend” the wires in the circuit diagram to better see the nodes like so:

Notice, as with the previous question, we simply bend the “empty” wire segments so that it is easier for us to visualize. This diagram is still the same as the original diagram.

Now, we will label our nodes and currents.

Looking at the orange node, V_1, we can write a KCL equation.

Let us express these currents in terms of voltage and resistance using I=\dfrac{V}{R}.

\dfrac{V_1-V_2}{4k}+\dfrac{V_1}{2k}+2m=4m\,\,\,\color{orange} {\text{(eq.1)}}

Now, let us write a KCL equation for the purple node, V_2.

Expressing these in terms of voltage and resistance gives us:

\dfrac{V_2-V_1}{4k}+\dfrac{V_2}{4k}+\dfrac{V_2}{12k}+4m=6m\,\,\,\color{purple} {\text{(eq.2)}}

Solving equations 1 and 2 simultaneously gives us (see full steps):

V_2=5.33 v

From our diagram, we know:

I_0=\dfrac{V_2}{12k}

I_0=\dfrac{5.33}{12k}

I_0=0.44 mA

#### Final Answer: