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Fundamentals of Electric Circuits

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3.3 Nodal Analysis with Voltage Sources 91<br />

Find the node voltages in the circuit <strong>of</strong> Fig. 3.12. Example 3.4<br />

3 Ω<br />

+ v x −<br />

20 V<br />

3v<br />

6 Ω<br />

x<br />

2<br />

1 + − + − 4<br />

2 Ω 10 A<br />

4 Ω<br />

1 Ω<br />

Figure 3.12<br />

For Example 3.4.<br />

Solution:<br />

Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL<br />

to the two supernodes as in Fig. 3.13(a). At supernode 1-2,<br />

i 3 10 i 1 i 2<br />

Expressing this in terms <strong>of</strong> the node voltages,<br />

or<br />

At supernode 3-4,<br />

v 3 v 2<br />

6<br />

10 v 1 v 4<br />

3<br />

v 1<br />

2<br />

5v 1 v 2 v 3 2v 4 60<br />

(3.4.1)<br />

or<br />

i 1 i 3 i 4 i 5 1<br />

v 1 v 4<br />

3<br />

v 3 v 2<br />

6<br />

4v 1 2v 2 5v 3 16v 4 0<br />

v 4<br />

1 v 3<br />

4<br />

(3.4.2)<br />

3 Ω<br />

+ v<br />

i x −<br />

1<br />

i 1<br />

6 Ω<br />

v<br />

v 2 v 3 1 v 4<br />

i i<br />

i 3 3<br />

2 i 5 i 4<br />

2 Ω 10 A<br />

4 Ω 1 Ω<br />

+<br />

3 Ω<br />

v x −<br />

Loop 3<br />

20 V<br />

i 3<br />

3v x<br />

+<br />

+ −<br />

+ −<br />

+<br />

6 Ω<br />

+<br />

+<br />

v 1 Loop 1 v 2 v 3 Loop 2 v 4<br />

− − − −<br />

(a)<br />

Figure 3.13<br />

Applying: (a) KCL to the two supernodes, (b) KVL to the loops.<br />

(b)

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