Basic circuit math

Discussion in 'General Electronics Chat' started by Razor Concepts, Oct 22, 2012.

  1. Razor Concepts

    Thread Starter Senior Member

    Oct 7, 2008
    Here is the circuit, and I am trying to find V2:
    The current is conventional, and a constant 2A. The 8v voltage source is also constant.

    I try to think of these as how they would work in real life, but to me it just looks like the circuit will blow up, as it seems impossible to get a constant 2 amps on the left and a constant 8 volts on the right simultaneously.

    Any ideas?
  2. Razor Concepts

    Thread Starter Senior Member

    Oct 7, 2008
    I got the answer - don't treat the 8v as a constant-voltage source. V2 = 20 volts
  3. crutschow


    Mar 14, 2008
    Not true. You did treat the 8V as a constant-voltage source (since that is how it is defined). Thus 2A times 6Ω gives 12V which is added to the constant-voltage 8V, giving the 20V you came up with.
  4. electron_prince

    Active Member

    Sep 19, 2012
    Current flowing in the circuit is 2 A

    therefore V2 = 8 Volts + (6 * 2) Volts

    Edit: sorry i didn't read your question completely last time.

    If you plot the characteristic graph for voltage and current source (voltage on the x axis and current on y axis), then for voltage source it is vertical and for current source it is horizontal.

    Note: considered ideal voltage and current source.
    Last edited: Oct 23, 2012
  5. JMac3108

    Active Member

    Aug 16, 2010

    For this circuit you just have to think about the two sources and what they are doing.

    The voltage source is forcing 8V at the right side of the 6 ohm resistor.

    The current source is forcing 2A to flow through the entire loop. Yes, this means 2A back through the voltage source.

    Given this, you can see how others calculated the 20V at V2. The 6 ohm resistor has V=(2A)(6ohm)=12V across it. The 2A current is flowing left-to-right in your diagram, therefore the voltage at V2 is 12V above the 8V voltage source, or 20V.

    Your real-world comment is valid. In the real world the circuit might not work. It depends on the properties if the voltage and current source. Maybe the voltage source can sink current, and maybe it can't. But for purposes of this problem and circuit analysis, you have to treat them as ideal sources.
  6. BillB3857

    AAC Fanatic!

    Feb 28, 2009
    Maybe the voltage source is a wet cell battery of 4 cells at 8 volts along the charge curve.