I know that you can short out a voltage source by placing a wire in parallel with it so all the current the source drives goes to the wire instead of to another branch with a resistor.

But can you short out a current source? I never saw my professor do it.

For example, can you go from this:

enter image description here

To this:

enter image description here


  • 3
    \$\begingroup\$ Please use the circuit editor, this mess is ambiguous. \$\endgroup\$
    – pipe
    Sep 24, 2017 at 0:16

1 Answer 1


Certainly the current source in your first circuit is useless. Its terminals are the same node, so it does not interact with the rest of the circuit, just circulates current back to itself. You can delete it from your analysis — or split at that node into two independent circuits, one with the shorted current source and one with everything else — and get the same result.

However, this is not usually considered analogous to shorting out a voltage source. A shorted ideal voltage source is a contradiction, and a shorted non-ideal voltage source will either destroy itself, shut down, or reach a current limit which reduces its output voltage.

The analogue of this in a current source is an open circuit. If you disconnect one or both terminals of a current source (or break the circuit elsewhere), then it is impossible for the specified current to flow, so this is a contradiction with an ideal source, and a non-ideal source will reach a voltage limit (“compliance voltage”) and thereby fail to supply the specified current.

(And for completeness: the fourth case is a voltage source in an open circuit — which, like the shorted current source you drew, does nothing and can be deleted from the circuit.)

  • \$\begingroup\$ If you keep current source C connected, like that, will there be current flowing through itself? Around the empty wire and come back. \$\endgroup\$ Sep 24, 2017 at 0:10
  • \$\begingroup\$ @user132522 yes, current will flow through the current source if you keep c connected \$\endgroup\$
    – BeB00
    Sep 24, 2017 at 0:13
  • \$\begingroup\$ But all of that will only flow through the empty wire? \$\endgroup\$ Sep 24, 2017 at 0:13
  • \$\begingroup\$ @user132522 I've updated my answer to say more on that. \$\endgroup\$
    – Kevin Reid
    Sep 24, 2017 at 0:14
  • 3
    \$\begingroup\$ @user132522 At the level of circuit analysis, we aren't worrying about “how”, there are just nodes and components. Perhaps you should post a different question. \$\endgroup\$
    – Kevin Reid
    Sep 24, 2017 at 0:22

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