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.

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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
Expressing this in terms of the node voltages,
6v3−v2+10=3v1−v4+2v1
or
5v1+v2−v3−2v4=60 (3.4.1)
At supernode 3-4,
i1=i3+i4+i5⟹3v1−v4=6v3−v2+1v4+4v3
or
4v1+2v2−5v3−16v4=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=0⟹v1−v2=20 (3.4.3)
For loop 2,
−v3+3vx+v4=0
But vx=v1−v4 so that
3v1−v3−2v4=0 (3.4.4)
For loop 3,
vx−3vx+6i3−20=0
But 6i3=v3−v2 and vx=v1−v4. Hence
−2v1−v2+v3+2v4=20 (3.4.5)
We need four node voltages, v1,v2,v3, and v4, 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=v1−20. Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives
6v1−v3−2v4=80 (3.4.6)
and
6v1−5v3−16v4=40 (3.4.7)
Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as
⎣⎢⎡366−1−1−5−2−2−16⎦⎥⎤⎣⎢⎡v1v3v4⎦⎥⎤=⎣⎢⎡08040⎦⎥⎤
Using Cramer’s rule,
Δ=∣∣∣∣∣∣∣366−1−1−5−2−2−16∣∣∣∣∣∣∣=−18,Δ1=∣∣∣∣∣∣∣08040−1−1−5−2−2−16∣∣∣∣∣∣∣=−480
Δ3=∣∣∣∣∣∣∣36608040−2−2−16∣∣∣∣∣∣∣=−3120,Δ4=∣∣∣∣∣∣∣366−1−1−508040∣∣∣∣∣∣∣=840
Thus, we arrive at the node voltages as
v1=ΔΔ1=−18−480=26.667 V,v3=ΔΔ3=−18−3120=173.333 Vv4=ΔΔ4=−18840=−46.667 V
and v2=v1−20=6.667 V. We have not used Eq. (3.4.5); it can be used to cross check results.
