Series DC Circuits Worksheet Qs #9-10

Discussion in 'Homework Help' started by Circuits123, Feb 6, 2014.

  1. Circuits123

    Thread Starter New Member

    Dec 7, 2012
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    You pick the reference point.

    If you choose from left to right, then you have: +3-6-4.5
    If you choose from right to left, then you have: +4.5+6-3
  3. Circuits123

    Thread Starter New Member

    Dec 7, 2012
    Ok. I think I see that. But is there a way to describe it as a rule? It seems like you take the sign of the total voltage and apply it to the side/reference point you started from. Is that it?
  4. tshuck

    Well-Known Member

    Oct 18, 2012
    What do you mean? The polarity is shown.

    The (ideal)voltage sources guarantee that the positive terminal is X volts above the negative terminal.
  5. WBahn


    Mar 31, 2012
    But the questions are asking for the voltage across the bulb and that polarity is not given.

    What (I'm pretty sure) you are expected to do is determine the absolute value of the voltage across the bulb and then to identify which side of the bulb is at the higher voltage and put a '+' sign there.

    The most systematic way to do it is to just pick one side of the bulb as your reference and put a '-' sign there. Then go around the circuit to the other side of the bulb adding up the voltage gains along the way. When you get to the other side you will either have a positive number or a negative number. If you have a positive number, then the side you finished at is more positive than the side you chose as a reference, so you can put your '+' sign where you ended up. If the number is negative, then you guessed wrong and the side you started at is the more positive side and so turn the original '-' sign into a '+' sign.

    And note that it would be perfectly legitimate to just stick with your initial pick and say that the voltage across the bulb is -16V and be done with it -- though that is not what is being asked for in these particular problems.