Question 3.4: Find the node voltages in the circuit of Fig. 3.12.

Find the node voltages in the circuit of Fig. 3.12.

Screenshot 2022-06-15 113243
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Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to the two supernodes as in Fig. 3.13(a). At supernode 1-2,

i3+10=i1+i2 i_{3}+10=i_{1}+i_{2}

Expressing this in terms of the node voltages,

v3v26+10=v1v43+v12 \frac{v_{3}-v_{2}}{6}+10=\frac{v_{1}-v_{4}}{3}+\frac{v_{1}}{2}

or

5v1+v2v32v4=60 5 v_{1}+v_{2}-v_{3}-2 v_{4}=60                (3.4.1)

At supernode 3-4,

i1=i3+i4+i5v1v43=v3v26+v41+v34 i_{1}=i_{3}+i_{4}+i_{5} \quad \Longrightarrow \quad \frac{v_{1}-v_{4}}{3}=\frac{v_{3}-v_{2}}{6}+\frac{v_{4}}{1}+\frac{v_{3}}{4}

or

4v1+2v25v316v4=0 4 v_{1}+2 v_{2}-5 v_{3}-16 v_{4}=0                   (3.4.2)

We now apply KVL to the branches involving the voltage sources as shown in Fig. 3.13(b). For loop 1,

v1+20+v2=0v1v2=20 -v_{1}+20+v_{2}=0 \quad \Longrightarrow \quad v_{1}-v_{2}=20                     (3.4.3)

For loop 2,

v3+3vx+v4=0 -v_{3}+3 v_{x}+v_{4}=0

But vx=v1v4 v_{x}=v_{1}-v_{4} so that

3v1v32v4=0 3 v_{1}-v_{3}-2 v_{4}=0                   (3.4.4)

For loop 3,

vx3vx+6i320=0 v_{x}-3 v_{x}+6 i_{3}-20=0

But 6i3=v3v2 and vx=v1v4 6 i_{3}=v_{3}-v_{2} \text { and } v_{x}=v_{1}-v_{4} . Hence

2v1v2+v3+2v4=20 -2 v_{1}-v_{2}+v_{3}+2 v_{4}=20                     (3.4.5)

We need four node voltages, v1,v2,v3, and v4 v_{1}, \quad v_{2}, v_{3}, \text { and } v_{4}, and it requires only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth equation is redundant, it can be used to check results. We can eliminate one node voltage so that we solve three simultaneous equations instead of four. From Eq. (3.4.3), v2=v120 v_{2}=v_{1}-20 . Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives

6v1v32v4=80 6 v_{1}-v_{3}-2 v_{4}=80                     (3.4.6)

and

6v15v316v4=40 6 v_{1}-5 v_{3}-16 v_{4}=40                       (3.4.7)

Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as

[3126126516][v1v3v4]=[08040]\left[\begin{array}{rrr}3 & -1 & -2 \\6 & -1 & -2 \\6 & -5 & -16\end{array}\right]\left[\begin{array}{l}v_{1} \\v_{3} \\v_{4}\end{array}\right]=\left[\begin{array}{r}0 \\80 \\40\end{array}\right]

Using Cramer’s rule,

Δ=3126126516=18,Δ1=012801240516=480\Delta=\left|\begin{array}{rrr}3 & -1 & -2 \\6 & -1 & -2 \\6 & -5 & -16\end{array}\right|=-18, \quad \Delta_{1}=\left|\begin{array}{rrr}0 & -1 & -2 \\80 & -1 & -2 \\40 & -5 & -16\end{array}\right|=-480

Δ3=302680264016=3120,Δ4=31061806540=840\Delta_{3}=\left|\begin{array}{rrr}3 & 0 & -2 \\6 & 80 & -2 \\6 & 40 & -16\end{array}\right|=-3120, \quad \Delta_{4}=\left|\begin{array}{rrr}3 & -1 & 0 \\6 & -1 & 80 \\6 & -5 & 40\end{array}\right|=840

Thus, we arrive at the node voltages as

v1=Δ1Δ=48018=26.667 V,v3=Δ3Δ=312018=173.333 Vv4=Δ4Δ=84018=46.667 V\begin{gathered}v_{1}=\frac{\Delta_{1}}{\Delta}=\frac{-480}{-18}=26.667  V , \quad v_{3}=\frac{\Delta_{3}}{\Delta}=\frac{-3120}{-18}=173.333  V \\v_{4}=\frac{\Delta_{4}}{\Delta}=\frac{840}{-18}=-46.667  V\end{gathered}

and v2=v120=6.667 V v_{2}=v_{1}-20=6.667  V . We have not used Eq. (3.4.5); it can be used to cross check results.

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