Series Current and Voltage source

Discussion in 'Homework Help' started by afaik, Mar 29, 2009.

  1. afaik

    Thread Starter Member

    Nov 2, 2008

    This question asks for the voltage across the current source and the voltage across the load.

    The given answer is:

    But I'm not sure how exactly to properly apply the rules.

    The book explains how to handle the circuit with a single source and load but not with different sources and a load. Sorry for such an elementary problem but it's got me scratching my head long enough I decided to ask the pros.

    A side question that will help clear things up for me: Electrons leave from the negative terminal and flow to the positive terminal with the voltage source, but what about the current source? Does it's direction refer to conventional or electron current?
  2. thatoneguy


    Feb 19, 2009
    It is a series circuit, so all currents are the same. So with 2mA flowing through the 7k resistor a 14V drop is created from ground.

    The voltage across the resistor will be V_R = 2mA \cdot 7k \Omega = -14V

    The voltage across the current source will be V_s=-14V-6V = -20V

    Since there isn't enough voltage in the circuit to have another negative voltage and still have the loop add up to zero, the 20V must be positive, i.e. an initial assumption of direction was incorrect.
  3. mik3

    Senior Member

    Feb 4, 2008
    Another way to think of that is that the current source (ideal) has infinite impedance and thus the voltage source has no effect on the current through the circuit. Also, an ideal voltage source has zero internal impedance and you can make the calculations Thatoneguy said.
  4. Ratch

    New Member

    Mar 20, 2007

    OK, we have a current source and a voltage source fighting each other. Since the tails of the arrows conververge at a common node, lets consider the junction of the current source and resistor to be "ground". The current source is going to supply the resistor with 14 volts so 2 ma goes through it. It is going to overcome a 6 volt back voltage from the voltage source, thereby increasing its required voltage output to 20 volts as the answer says. Shame on whoever drew the diagram with both voltages marked "V" instead of V1 and V2, and no designated common measuring point.

    Did you look elsewhere in this forum for an answer, especially a post only a few days old? . You use the industry standard of conventional current flow. All semiconductors are marked that way. So is your voltmeter and ammeter. Who cares if the charge carriers are electrons (which they are in this case) or positrons? The values of voltage and current will be the same.

    So lets look at the circuit both ways. Conventional current assumes positive charges come from the left side of the current source and the left side is 20 volts positive. The battery drops it down to 14 volts for the resistor.

    Now the "real" direction of the current is from the opposite right side. It goes through the resistor and makes 14 volts across the resistor. Same polarity too. That is because negative charges going in one direction is the same as positive charges going in the opposite direction. So don't get hung up what polarity the charge carriers are. Always use the industry standard conventional charge flow and evaluate the "real" flow on a case by case basis if you really have to know.