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I'd like to help a fellow student in getting the ins and outs of Kirchhoff's current and voltage law. Thing is, it's been a while for me...

I remember that the current law dictates the sum into and out of a node to be zero, and the sum of the voltage inside a loop again zero. I may also assume the initial current directions.

The circuit in question has 10 resistors and five ideal voltage sources.

enter image description here I am not sure if I may do that, but IIRC, I can combine those resistors that are in series (same current). That is, R1, R2 and R3 become R123 (3k), the same goes for R7, R8, R9 and R10 (3,45k).

This yields I1 - I5 as the unknowns so I need a system of five linear equations:

  • Loop 1: 0 = V1 - I1 * (R1 + R2 + R3) - I2 * R5 - V2
  • Loop 2: 0 = V2 + I2 * R5 + V5 - I4 * R6 - V3 - I3 * R4
  • Loop 3: 0 = V3 + I4 * R6 - I5 * (R7 + R8 + R9 + R10) + V4
  • Node 1: 0 = I1 - I2 - I3
  • Node 2: 0 = I3 - I4 - I5

    Plugging this into Mathemathica I get

    • I1 = 2.1 mA
    • I2 = 0.36 mA
    • I3 = 1.74 mA
    • I4 = -1.72 mA
    • I5 = 3.45 mA

I am not so much interested in the actual values but if my understanding is correct.

Thanks for reading! Chris

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  • \$\begingroup\$ Seems like you are on the right track. But just remember that if you combine resistors in series, you cause a Voltage node to disappear. For example, if the question was to find the voltage from V1+ terminal to V5+ terminal, then combining R1 with R2 and R3 would be problematic because it would cause one of your voltage nodes to totally disappear. In practice it is OK as long as you don't forget what the original circuit looked like (before you combined things). \$\endgroup\$ – mkeith May 4 '16 at 4:06
  • \$\begingroup\$ Thank you for the hint! Confusingly, their instructor gave a different solution with different equations, and then only a partial solution. At this point, I'd simulate that whole thing but the circuit lacks an indication of where GND is, so PSpice won't let me... \$\endgroup\$ – user2286339 May 4 '16 at 20:23
  • \$\begingroup\$ Pick any node as GND. Maybe the minus side of V5. Then you can at least solve for the currents. \$\endgroup\$ – mkeith May 5 '16 at 4:03
  • \$\begingroup\$ This is good to know. I had thought of GND as changing everything depending on its location. The currents that PSpice calculates now (I1 = 1.991mA, I2 = 514uA, I3 = 1.476mA, I4 = -1.148mA and I5 = 2.624mA) are nowhere near my own solution so obviously I made a mistake in the five equations, but where ? \$\endgroup\$ – user2286339 May 6 '16 at 17:06
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    \$\begingroup\$ I finally found out that I had to use one additional node equation. General rule: If there are 'n' nodes, we need n - 1 node equations. If there are 'b' branches, we need b - (n - 1) mesh equations. In the circuit above, n = 4 and b = 6 so this requires 3 node and 3 mesh equations. Source: en.wikipedia.org/wiki/Network_analysis_(electrical_circuits). Higly recommended article anyway. \$\endgroup\$ – user2286339 May 8 '16 at 21:47
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[Copy of inline comment]

I finally found out that I had to use one additional node equation.

General rule: If there are 'n' nodes, we need n - 1 node equations. If there are 'b' branches, we need b - (n - 1) mesh equations. In the circuit above, n = 4 and b = 6 so this requires 3 node and 3 mesh equations.

Source: en.wikipedia.org/wiki/Network_analysis_(electrical_circuits). Higly recommended article anyway.

Chris

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