Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Archived Quick circuit help/confirmation

  1. Oct 19, 2010 #1
    1. The problem statement, all variables and given/known data

    This should be a really easy circuit, but I need to see if I have it done right.

    1V source, 1A source and a 1 Ohm resistor in parallel (forgot to write the 1 Ohm resistor on the image.)

    Find the power of each element.

    2. Relevant equations

    Ohm's law, KCL, KVL

    3. The attempt at a solution


    this just seemed kinda counter intuitive since it appears that there is no current running through the 1v source. Any help would be greatly appreciated.
  2. jcsd
  3. Feb 5, 2016 #2


    User Avatar

    Staff: Mentor

    The original image has disappeared, so here's one based on the text description:

    Currents have been indicated in anticipation of calculating the power associated with the components.

    The voltage source will hold the potential across the 1 Ω resistor at 1 V. Ohm's law tells us that the current ##i_R## must be 1 Amp.

    The current supply is pushing 1 Amp into the circuit, and this will satisfy the requirements of the 1 Amp through the resistor.

    This leaves the voltage supply to consider. Note that it does not have to supply or sink any current in order for the potential across the resistor to remain at 1 V, since the current source is taking care of that. The voltage source could be removed from the circuit and the voltage drop across the resistor would remain at 1 V.

    Now for the power produced or consumed by the components. Since the voltage supply is not passing any current there is no power associated with it. It neither produces nor consumes any power.

    The current supply is producing 1 Amp at 1 Volt, so it is producing 1 Watt of power to the circuit.

    The resistor is carrying 1 Amp of current and producing a 1 Volt potential drop in the direction of the current flow, so it is dissipating 1 Watt of power (the same power that the current source is injecting into the circuit).

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?