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Homework Help: Grounded circuit, 2 batteries and 2 resistors

  1. Mar 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Picture a circuit grounded at the left lower corner. Then as you go up the left side of the circuit, you have a 9V battery. The top left corner is labeled a. Then on the top of the circuit, you have a 2 ohms resistor. Then on the right side of the circuit, you have a 6V battery. The right lower corner is labeled b. Then on the bottom side of the circuit you have a 1 ohm resistor.
    What is the value of the potential at points a and b?

    2. Relevant equations
    Kirchhoff's loop law
    I = Epsilon/R
    Delta Vr = - I*R

    3. The attempt at a solution

    First Kirchhoff's loop law to find the current in the circuit. I assumed the clockwise current was the same in both resistors.

    --> 9V - 2 I - 6V - 1 I = 0
    3V = 3I --> I = 1A

    I know the potential at point a is 9V. The potential at the lower left corner, where the circuit is grounded, is zero. As you go up the circuit, you have gained 9 V by "passing through" the 9V battery.

    Then you pass through the top resistor.
    The potential difference across that resistor is given by delta Vr = - I * R = -1A*2omhs =-2V. So by the time you reach the right top corner, you are at 9V-2V= 7V.

    Then you pass through the 6V battery. You go from the positive terminal to the negative terminal, so you lose 6V. Hence, at point b you are 7V-6V= 1V.

    Is my reasoning any good?
  2. jcsd
  3. Mar 31, 2016 #2


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    What about polaritiy of the batteries? A diagram would be helpful.
  4. Mar 31, 2016 #3
  5. Mar 31, 2016 #4


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    Yes, that all works.
  6. Mar 31, 2016 #5
    Muchas gracias! !
  7. Apr 1, 2016 #6


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    You can also get to b by going "up" through the 1 Ohm....

    0 + (1*1) = 1V
  8. Apr 1, 2016 #7
    True that.
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