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enter image description here

My book above uses nodal analysis to find the current and voltage across the resistors.

Now consider this alternate circuit:

enter image description here

Look at the branch that doesn't any components (the second straight line with no components). How would I find the current across it? I only know Kirchoff's Laws and Nodal Analysis so far. Can the current across it be found using these techniques?

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  • \$\begingroup\$ Are you sure you copied the 2nd circuit correctly? \$\endgroup\$
    – Spehro Pefhany
    Mar 18, 2014 at 1:54
  • \$\begingroup\$ @SpehroPefhany I didn't copy it. I modified the first circuit because I was curious how the modified circuit would act. Is there a problem with it? \$\endgroup\$
    – dfg
    Mar 18, 2014 at 2:02
  • \$\begingroup\$ Think about the wire and the 6kOhm resistor as a current divider, with the wire having a resistance of 0Ohms. What fraction of the incoming current goes through the resistor, and what fraction goes through the wire? \$\endgroup\$
    – Shamtam
    Mar 18, 2014 at 2:05
  • \$\begingroup\$ The 6k resistor has no voltage across it, so it cannot have the 2.5mA current flow shown. \$\endgroup\$
    – Spehro Pefhany
    Mar 18, 2014 at 2:05
  • \$\begingroup\$ @Shamtam Sorry, I typed up my answer and posted it, not realizing that you had commented in the mean-time... \$\endgroup\$
    – bitsmack
    Mar 18, 2014 at 2:07

1 Answer 1

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In this case, you first want to reduce the circuit. I'm going to have to explain this without pictures, sorry. But I think it'll be pretty clear anyway!

  1. The first thing to do is to ignore the small arrow marks that are showing the branch currents - they are no longer accurate once the circuit changed. Neither are V1 and V2.

  2. Look at the 6k resistor. It has effectively been removed from the circuit. That is, there will be no current through this resistor. There are a few ways to see why this is so. Here are two of them:

    • The resistor is in parallel with a zero-Ohm path. Any current flow will happen through this short, and nothing will flow through the more difficult path of the resistor.

    • There can be no voltage that exists across the short. Therefore, no voltage will be presented across the 6k resistor, either, since they are in parallel.

  3. So, now, you are left with the 1mA source, flowing though a 6k resistor in parallel with 12k resistor. Easy to analyse with either of the techniques you mentioned.

Good luck :)

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