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Power in circuits

  1. Dec 4, 2006 #1
    Here's the question:

    As seen in the figure below, R1 = 4.00 W, R2 = 8.00 W, R3 = 8.00 W, R4 = 4.00 W, E1 = 50.0 V and E2 = 20.0 V.

    (I can't copy image so here it is)

    ...........R1
    _____/\/\/\__________________
    |...................|........|.............|
    |...................|........|.............|
    __..................|........|............._
    _..E1..............\......./.R3.........__..E2
    |...............R2./.......\..............|
    |...................|........|.....R4.....|
    |____________|_____|__/\/\___|

    (Current from E1 flows to R1 and current from E2 flows to R4.)

    I'm supposed to find the power supplied to R1, R2, R3 and R4.

    I chose to find the power supplied to R4 first

    The 50V charge encounters a resistance of 8 ohm (4+ (64/16)) before reaching R4 and the 20V charge encounters no resistance before reaching R4.

    First I tried not including the resistance in R4 which would give
    (50/8)^2 * 4 + 20^2 * 4 and that wasn't correct.

    Then I tried including the resistance in R4 which would give (50/12)^2 * 4 +
    (20/4)^2 * 4 and that wasn't correct either.

    I also tried combining the currents before squaring them -
    (50/8+20)^2 * 4 and (50/12+20/4)^2 * 4 - and that did not give me the right answer.

    I don't know what else to try and the book/internet isn't much help. Any ideas would be appreciated.

    Thanks a lot.
     
    Last edited: Dec 4, 2006
  2. jcsd
  3. Dec 4, 2006 #2
    I'm fairly certain the direction of the current is as specified. I used the loop rule on the outside loop and it seems to be correct.
     
    Last edited: Dec 4, 2006
  4. Dec 4, 2006 #3

    berkeman

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    Staff: Mentor

    You need to solve for the node voltages first, and then use those node voltages to give you the powers.

    Call the bottom of E1 ground, then solve for the three unknown voltages by writing their three node equations (the sum of the currents out of a node must equal zero). Please post that work, and then we can check the powers you calculate at the end.
     
  5. Dec 4, 2006 #4
    I'm not sure if this is what you mean but I think I forgot the negative charge from both terminals.

    So for R1 the currents would be 50/4 from E1+, 20/12 from E2+ and -20/8 from E2-.

    For R4 the currents would be 20/4 from E2+, 50/12 from E1+ and -50/8 from E1-.

    Now I'm not sure how to find the current to R2 or R3.
    For R3: Is it 50/8 from E1+, -50/4 for E1- and 20/6 for E2+ and -20/2 for E2-?
    20/6 for E2+ because it runs through R4 and one half of the parallel series and -20/2 for E2- because it runs through only one half of the parallel series.

    If this is right then the current to R2 is:
    50/6 from E1+, -50/2 from E1-, 20/8 from E2+ and -20/4 for E2-.

    Also, should I add up the currents and then square them or square them seperately? Thanks again.
     
  6. Dec 4, 2006 #5

    berkeman

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    Staff: Mentor

    Are you familiar with how to use the KVL and KCL equations to solve for voltages and currents in circuits? Check out the "Analysis of Resistive Circuits" link near the bottom of this wikipedia page:

    http://en.wikipedia.org/wiki/KVL
     
  7. Dec 4, 2006 #6
    Alright, I'm getting somewhere now but I'm not sure how many loops there are in this circuit. I think it's 5:

    Loop 1: E1 to R1 to R2 to E1
    Loop 2: E1 to R1 to R3 to E1
    Loop 3: E1 to R1 to E2 to E1
    Loop 4: E2 to R4 to R3 to E2
    Loop 5: E2 to R4 to R2 to E2

    Is this right?
     
  8. Dec 4, 2006 #7

    berkeman

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    Staff: Mentor

    I'm partial to the KCL, where you set up one equation per node (excluding ground). Can you set up the 3 equations for the nodes I mentioned in my post #3?
     
  9. Dec 4, 2006 #8
    Are the nodes at the resistors or at the intersections of the conducting wire? I checked out the wikipedia site but it's still not making a whole lot of sense to me.

    Thanks.

    Is it anything like (0-V1)/4 + (V1-V3)/8 + (V2-V4)/8 = 0A and V2 - V4 = 20V ?

    with V1, V2, V3, V4 going clockwise from the node between R1 and R2.
     
    Last edited: Dec 4, 2006
  10. Dec 4, 2006 #9

    berkeman

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    Staff: Mentor

    After looking again at the circuit, I'd organize it this way:

    call the bottom of E1 ground and add a ground symbol to remind you that it is at zero volts (0V). Then call the intersection of R1, R2 and R3 (and the - terminal of E2) a voltage Vx. Then write an equation that shows that the sum of the currents out of the Vx node must equal zero total. The equation looks something like this:

    [tex]\frac{V_x - E1}{R1} + \frac{V_x}{R2} + (two other terms) = 0[/tex]
     
  11. Dec 4, 2006 #10
    Vx/R3 + (Vx+E2)/R4 ?
     
  12. Dec 4, 2006 #11
    I get something different. Summing currents into the node of R1,R2,R3 (and turning E2 voltage around by making it negative... show below):

    ...........R1
    _____/\/\/\__________________
    |...................|........|.............|
    |...................|........|.............|
    __..................|........|.............__
    _..E1..............\......./.R3........._..(-E2)
    |...............R2./.......\..............|
    |...................|........|.....R4.....|
    |____________|_____|__/\/\___|
    ..................____....................
    ...................__...................... <-------(I also added a ground)
    ...................._.......................


    [tex]\frac{E_1- (-E_2)}{R_1} + \frac{0-(-E_2)}{R_2} +\frac{0-(-E_2)}{R3}=0[/tex]
     
  13. Dec 5, 2006 #12

    berkeman

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    Staff: Mentor

    Correct. Now set the whole sum equal to zero and solve for Vx. That gives you all the voltages in the circuit, so the resistor powers are just obtained from

    [tex]P = \frac{V^2}{R}[/tex]

    Sorry, I don't understand Mindscrape's equation, but it may also be valid.
     
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