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Homework Help: Square circuit

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    In the circuit given (attached) , point S is earthed , what are the electrical potentials at point P , Q and R .

    2. Relevant equations

    Kirchoff's second law .

    3. The attempt at a solution

    The electric potential at point S is 0 . By applying Kirchoff's second law ,



    i am not sure where to go from here .

    Attached Files:

  2. jcsd
  3. Mar 28, 2010 #2

    Doc Al

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    Staff: Mentor

    You found the current. In what direction does it flow? Use that to find the voltage drops across each resistor.
  4. Mar 31, 2010 #3
    ok , the current is flowing in the clockwise direction . Could you elaborate a little further on the voltage calculation of each resistors ?

    But the question is asking for the potential at points P,Q and R respectively . How is it related to the potential difference across the resistors .

    Thanks !
  5. Mar 31, 2010 #4

    Doc Al

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    Staff: Mentor

    Sure. The voltage across each resistor is given by Ohm's law: ΔV = IR. To find the sign of the voltage across the resistor, you need to know the direction of the current. Hint: Does current across a resistor flow from lower to higher potential or from higher to lower?

    To find the potential at any point with respect to some reference, just add up the voltage drops across each piece. Here your reference is point S, which is marked as ground and thus is at 0 Volts. To find the potential of point P, for example, you need to find ΔV from S to P. S to P contains a battery, so what's ΔV? To find the potential of point Q, you'll add up the voltage drops across S-P and P-Q. What's the voltage drop from P to Q across that 2 Ω resistor?
  6. Mar 31, 2010 #5
    thanks ! Now , i understand this question better . Let me attempt again .

    Since the emf of the 15 V battery > emf of 5V battery , the current will be flowing in the clockwise direction .

    pd across sp would simply be -5V .

    pd across 2 ohm resistor is 2/(2+3) x (5-15)=-4V so pd across SQ is -5-4=-9V

    pd across 3 ohm resistor is 3/(2+3) x (5-15)=-6V so pd across SR is -15 V

    Are my reasonings correct ?
  7. Mar 31, 2010 #6

    Doc Al

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    Staff: Mentor


    Just for fun, realize that you can find the potential of a point (Q, say) by starting from S and going clockwise around the circuit or by going counter-clockwise. Do it both ways and check that you get the same answer.
  8. Mar 31, 2010 #7
    yay! Thanks ! Yeah , i have experimented with that
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