Solution

First, label each mesh (window pane) with a mesh current. For consistency, make each mesh in a clock-wise direction.

Now write KVL equations for each loop.

KVL for i1:

-18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0

then gather terms:

10i1 - 5i2 - 4 i3 -18V = 0

Note that the i1 term is positive, and all other current terms are negative (because they are all clockwise, all other panes will contribute a negative term). Let's do the other two panes with terms gathered up directly (write the total resistance of the loop multiplied by the mesh current that goes through that total resistance):

KVL for i2:

-5i1 + 10i2 - 3i3 - 12 = 0

KVL for i3:

-4i1 -3i2 +9i3 = 0

Now solve the three equations in three unknowns:

i1 = 7.02A
i2 = 6.28A
i3 = 5.21A

The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current through the 18V is i1, or 7.02A. All the branch currents are shown below from a Pspice simulation:

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© Calvin College, 2000
This page was written and is maintained by Steve VanderLeest.
It was last modified on 17-Feb-2003.