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When I try to solve this problem I get three equations:

  1. VR3+VR5=20
  2. VR1+VR3=-20
  3. VR1-VR5=-40

The problem is that you can obtain 1. by subtracting 3. from 2. so you actually have only two equations with three variables. What am I missing?

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Someone on Quora helped me out. I didn't think to apply the current law i.e. IR1+IR5=IR3. Then it's easy to solve.

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  • \$\begingroup\$ If you write out the easy solution here, I'll +1 your answer and your question!! \$\endgroup\$ – jonk Apr 19 '20 at 2:51
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Now write those resistor voltages in terms of the mesh currents.

For example, if we take the left-hand loop as mesh 1, and it's current to be going clockwise, then $$V_{R5}=I_1 R_5$$

Once you do that you'll have expressed your three resistor voltages in terms of only two mesh currents, and you'll be down to two unknowns for two equations.

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My problem was the opposite. It seems that KVL and KCL always give too many equations!

Any circuit can be modeled by a simple biconnected graph. A graph is made of faces, vertices and edges. According Euler's Characteristic, F + V = E + 2. The KVL gives one equation per face, and the KCL gives one equation per vertex. There is one (unknown) current per edge. Therefore you always have two more equations than you need.

The reason is that one KCL equation and one KVL equation is always degenerate (i.e. it is implied by the other equations). Therefore, when you are gathering your equations, just throw away one of each. It doesn't matter which one.

The OP circuit has F=3, V=2, and E=3. The OP correctly found three KVL equations (one per face) and observed the rank is only 2, so you need to get another equation from the KCL.

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