Solution

First, select a reference node and label the other nodes. All the nodes have three connected branches, so we will simply choose the bottom node as the reference.

Now write KCL equations for each node except the reference, in terms of the node voltages:

KCL at V1:

-3A + (V1-V2)/5 + (V1-V3)/1 = 0

KCL at V2:

(V2-V1)/5 + V2/3 + (V2-V3)/2 =0

KCL at V3:

(V3-V2)/2 + (V3-V1)/1 - 8A = 0

Now gather terms and clear up the fractions:

6V1 - V2 - 5V3 = 15

-6V1 + 31V2 - 15V3 =0

-2V1 -V2 +3 V3 = 16

Finally, solve the 3 equations in 3 unknowns.

V1 = 48.625V

V2 = 33 V

V3 = 48.75V

The current through the 5 ohm resistor can be found by Ohm's law:

I = (V1 - V2)/5 = 3.125A

The voltage over the 3A source is simply V1, or 48.625V.

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© Calvin College, 2003
This page was written and is maintained by Steve VanderLeest.
It was last modified on 17-Feb-2003 .