Use nodal analysis to find both V1 and Vo


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

Use nodal analysis to find both V1 and Vo

Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.

Solution:

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

Use nodal analysis to find both V1 and Vo

When writing equations, k=10^3 and m=10^{-3}

Note that we will use I_4 to calculate V_0.

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

I_1+I_2+2m=12m

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.

I_3+I_4+I_5=2m

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_1=\dfrac{252}{11} v

V_2=\dfrac{96}{11} v

 

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

I_4=\dfrac{V_2}{2k+1k}

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_1=\dfrac{252}{11} v

V_0=2.91 v

 

This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 3, question 3.6.

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