Label each mesh (window pane) with a mesh current. Then write the KVL equations for each pane. Note that we were forced to label the voltage over the current source (Vx) in order to write the voltage term there:
We now have an extra unknown (Vx), so we need another equation. It is found be relating the two mesh currents to the current source.
Note that i1 is positive because it is in the same direction of the source. I2 is negative because it is in the opposite direction as the source.
Now solve the three equations in three unknowns. I1 is found to be -320mA. Since ix is in the opposite direction of i1, then ix = 320mA.
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© Calvin College, 2004
This page was written and is maintained by Steve VanderLeest.
It was last modified on 17-mar-04 .