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The expressions

\$N-1=\# KCL \$

and

\$E-N+1=\# KVL \$

(where \$N\$=number of nodes; \$E\$=number of elements on the circuits) gives us the linear independent equations on a circuit.

But sometimes the expressions gives me, for example, 5, and I only have 3 unknowns. So, why is this wrong? What do the value of the expressions tell me? The independent equations I can write, or the independent equations I will need to analyze the circuit?

What about nodes? Are they the theoretically nodes or the practical nodes? I always consider on this expressions the theoretical nodes. I mean, it doesn't necessarily have to be a current divider, it can simply connects 2 elements of the circuit.

These equations aren't very accurate to me, and I would like you to help me figuring it out why.

Per example:

Here I have 8 nodes and 10 elements, so I got 7 independent KCL equations and 3 independent KVL equations, but I only want to know I,I1,I2,I3,I4,I5, so I just have 6 unknowns. Why I have more equations that unknowns? How can I select the 7 KCL equations in order to don't get dependent equations?

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  • \$\begingroup\$ You should define your terms. What are "N" and "E"? \$\endgroup\$ – The Photon Apr 11 '17 at 21:01
  • \$\begingroup\$ You should post the example it's giving you the wrong result to be sure. In general, the formulae give the number of linear independent equation required to solve for all voltages and currents in the linear circuit. Nodes are 'topological', in the sense that you have to collapse all joints to the same part of the circuit into a single node. For example, the whole Vcc rail will be compressed in a single node; the whole ground rail will be compressed in a single node. \$\endgroup\$ – Sredni Vashtar Apr 11 '17 at 21:19
  • \$\begingroup\$ @SredniVashtar I inserted an example \$\endgroup\$ – Vitor Aguiar Apr 11 '17 at 23:36
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The goal of circuit analysis is to find the currents through and the voltages across all the components. Ideal sources are a bit special because we don't have rules that relate the voltage the current. If you solve both KVL and KCL you will get all the voltages and currents. For KCL one node is a reference so N-1 is correct for the number of equations. For KVL it is the number of loops that matters. If using KCL and you have a voltage source you will need to create a super node or write and additional equation to completely solve. Similarly if using KVL with a current source you you will need and additional equation. If you are doing this right you will not get an over specified set of equations. Typically both KCL and KVL are not done. You can pick the one that is easier depending on the circuit topology. For example if there is a current source and a bunch of resistors, it will probably be easier to use KCL and avoid the extra equation. This would give you all the voltages. Then you would use Ohms law to get the currents. This saves doing KVL.

your circuit edited

For the I1 loop;

-I1xR1 - V2A + V1 - (I1-I2)xR2 = 0

For the I2 loop;

-(I2-I1)xR2 - V1 -(I2-I3)xR6 - V2 - (I2-I3)xR3 = 0

If you write the equation for I3 and an extra equation for the voltage across the 2A current source that I've labeled V2A in the I1 loop, I'll check back and tell you if I agree. If you solve these four equations you will know all the currents. Then you can find the voltages using Ohm's law. Instead of Ohm's law you could solve the node equations but that would be much more work.

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  • \$\begingroup\$ I edited my question \$\endgroup\$ – Vitor Aguiar Apr 11 '17 at 23:37
  • \$\begingroup\$ You have to solve for all to get the 7 you want. Sometime you may be able to use equivalent components to simplify things. For example if you use the series combination of R4 and R5 things would be more simple. And in this case if you use KCL you are going to add 3 equations for the voltage sources. If you use KVL you will need the 3 loop equations and an extra for the current source. In this case your best bet would be KVL because you only need to solve 4 equations and then Ohm's law to know everything. \$\endgroup\$ – owg60 Apr 12 '17 at 0:00
  • \$\begingroup\$ I don't understand. Suposedly I need 10 independent equations: 7 from KCL and 3 from KVL. I know how to get KVL. But what about KCL? Consider the circuit in the form it is now. \$\endgroup\$ – Vitor Aguiar Apr 12 '17 at 0:17

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