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Potential Difference between two points of a circuit

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Given the following circuit, what is Vab? Explain...


    2. Relevant equations
    Kirchoff's Law and Ohm's Law

    3. The attempt at a solution

    Here is my solution... Is this describing it correctly?

    Kirchoff’s Loop Law to find current:
    -10 Ω (I) + 30 Ω V – 100 Ω (I) – 5 Ω (I) + 15 V – 40 Ω (I) = 0
    45 V = 155 Ω (I)
    I = 45 V / 155 Ω or .29 Amps

    Potential Difference between points a and b:
    1. At point a, we will assume there to be zero potential.
    2. By going over the 5 Ω resistor, we lose 1.45 V (5 Ω * .29 A), thus leaving us at - 1.45 V.
    3. The 15 V power supply adds 15 V, leaving us with 13.55 V.
    4. The 40 Ω resistor takes away another 11.6 V (40 Ω * .29 A), leaving us with 1.95 V.
    5. The 5 Ω resistor that we come in contact with next doesn’t take anything away as it doesn’t have current running through it, so we’re left with 1.95 V.
    6. Now, we add another 10 V to the potential difference between the point a and point b due to the power supply.
    7. We now have a combined Potential Difference between point a and point b to be 11.95 V.
  2. jcsd
  3. Feb 27, 2007 #2
    I didnt check your math but the rest seems good to me.
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