Use nodal analysis to find both V1 and Vo in the circuit in Fig 3.6.

#### Solution:

Show me the final answerâ†“

Let us label the currents flowing from each node as follows:

Note that we will use I_4 to calculate V_0.

Let us write a KCL equation for node V_1, the orange node.

We can express these currents in terms of voltage and resistance using I=V/R.

\dfrac{V_1}{3k}+\dfrac{V_1-V_2}{6k}+2m=12m\,\,\,\color{orange} {\text{(eq.1)}}

We will now look at node V_2, which is the purple node and write a KCL equation.

Again, we will express these currents in terms of voltage and resistance.

\dfrac{V_2-V_1}{6k}+\dfrac{V_2}{2k+1k}+\dfrac{V_2}{6k}=2m\,\,\,\color{purple} {\text{(eq.2)}}(Remember that for I_4, the current that flows is across both resistors added together as they are in series)

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

V_2=\dfrac{96}{11} v

We will now use the value of V_2 to find V_0. Remember that

I_4=\dfrac{(\frac{96}{11})}{3k}

I_4=2.91 mA

Now,

V_0=(I_4)(1k)

V_0=(2.91m)(1k)

V_0=2.91 v

#### Final Answer:

V_0=2.91 v