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If I have a circuit where one branch contains a constant current source (e.g. 2A), does the current leading out of the current source equal the current leading into it (e.g. 4A) plus the current of the current source? That is, is the current leading out of the current source 2A+4A = 6A?

Or does the current source simply ensure that the current along the entire branch is equal to its value, e.g. if the constant current source was 2A the current along the entire branch would be 2A.

EDIT: Here is a diagram; I'm asking if \$I_A = I_B = 5A\$ or if \$I_B = I_A + 5A\$. Apologies for my poor MS paint skills currentSource.png

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  • \$\begingroup\$ if you think of 2 loops with a shared resistor in the middle, and both loops were "pushing" a constant current into the resistor, the current through the resistor (the shared element only!) would be the sum of these currents. A resistor in each loop only would only show the current from that loop present through it. \$\endgroup\$ – KyranF Jun 24 '15 at 21:50
  • \$\begingroup\$ Hi Kyran, I've added a diagram to clarify my question as I don't think I explained the source of my confusion properly. \$\endgroup\$ – NewDogOldTricks Jun 24 '15 at 22:00
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    \$\begingroup\$ What does Kirchoff's current law say about this? \$\endgroup\$ – The Photon Jun 24 '15 at 22:04
  • \$\begingroup\$ It means that the current through the current generator is constant and at the rated value, in your diagram Ia=Ib=5A. This means that both the input and output currents are 5A. For any 2-terminal component, input current=output current. \$\endgroup\$ – Chu Jun 24 '15 at 22:13
  • \$\begingroup\$ @ThePhoton Assuming the constant current source is not "adding" current, then I think both \$I_A\$ and \$I_B\$ are 5A according to KCL. Is this interpretation that it is not adding current correct? \$\endgroup\$ – NewDogOldTricks Jun 24 '15 at 22:15
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IA=IB=5A

Yes. The current entering a node has to be same as current leaving the node. Even if another current source of 5 A is connected in series, the equivalent current will be 5 A. This is similar to two similar voltage sources connected in parallel.

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Actual application of adding currents with current sources.

schematic

simulate this circuit – Schematic created using CircuitLab

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Current sources must be connected in // in order to work correctly and the load must be also //, otherwise they are fighting against each other. If connected in // then the currents sum up.

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