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Parallel circuit multiple choice question

  1. Apr 26, 2013 #1
    This past Monday I took an exam and got a 99, woot woot! But I can't seem to understand why I got the following question wrong:

    Complete the following statement: A simple series circuit contains a resistance R and an ideal battery. If a second resistor is connected in parallel with R,
    A) the voltage across R will decrease.
    B) the current through R will decrease.
    C) the total current through the battery will increase.
    D) the rate of energy dissipation in R will increase.
    E) the equivalent resistance of the circuit will increase.

    I picked B since I used the following reseasoning: Itotal = I1 + I2, therefore the current in R is decreased. My professor said that C is the correct answer. Although she tried to explain it, I couldn't understand it. Do you mind explaining it to me, please?

    Thank you so much!
     
  2. jcsd
  3. Apr 26, 2013 #2

    Danger

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    Gold Member

    This is not in my area at all, and I probably would have scored only "1" on that test (if "A" is the correct answer.) My knowledge of electrical stuff is almost non-existent. Other than adding a few outlets to my house and rebuilding a bunch of things like VCR's, my experience is nil. The only reason that I think of "A" is that I once had to run a 6VDC motor from a 12VDC power supply. A friend who knows about this kind of stuff told me to parallel wire a resistor with the same rating as the load of the motor to cut the voltage in half. He called it a "voltage divider". While I can't say for sure why, it worked.
     
  4. Apr 26, 2013 #3

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Your professor is correct. As long as the battery is an ideal battery (no internal source resistance), the output voltage is indepent of output current. So the voltage across the parallel combination of R//R is the same as it was across the initial single R. And since I=V/R, that means that the current through the first resistor will not change when the second R is put in parallel with it. All that happens is that twice as much current is drawn from the battery when you put the 2nd R in parallel with the first.

    Make sense now?
     
  5. Apr 26, 2013 #4

    CWatters

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    I can see where you went wrong...

    Your answer would be correct if an ideal battery was an current source. However an ideal battery is a voltage source.

    For a constant voltage source the current Itotal isn't constant it depends on the load...

    Itotal = Vbat/R1 + Vbat/R2
     
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