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Homework Help: Almost Solved my ele. hmwk, need someone to check it!

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    1) a 12m length of wire with uniform cross sectional area 2.75x10^-7 has a potential difference of 15v applied across it, producing a current of 20A
    a) find the resistivity of the wire B) what is the power dissipated in the wire?

    2) 2 circuits are made. In circuit 1, 4 resistors (1,2,3,6 ohms) are connected in series. in circuit 2, the same 4 resistors are connected in parallel. The battery emf is 24 volts in both cases. Calculate the equivalent resistance of the each circuit and the current through each resistor.

    determine the equivalent resistance and the total power deliverd by the battery for this ciruit.
    Click here if you cant see the pic

    3. The attempt at a solution
    1) a) r=I/V r=15/20 r=.75
    b) p=IV p=20*15 p=300

    2) Equivalent resistance for the series would be 12 because i just added the resistors up and i solved for current doing 24=I(12) V=IR so then the current through each resistor would be 2,4,6,12 respectively
    for the parallel i did 1/rp=1/r1+1/r2+1/r3+1/r4 and got 1/3=rp, i solved current using v*r=I
    24(1/3)=8 and then for each resistor the current was 24,12,8,4 respectively.

    3) I am totally confused on this one?
    Last edited: Feb 21, 2009
  2. jcsd
  3. Feb 21, 2009 #2
    1 a) is correct answer, but you have symbols for current and voltage wrong.

    b) There is an equation for resistivity, that includes lenght and cross sectional area.

    2) Req for series is correct, but the currents aren't. The current of one loop is constant inside that loop. So the currents through each resistor would be?

    Req for parralleled is wrong. Your method of finding the total current is right, but for each resistor it's wrong. Kirchof's law states that if you divide one wire into multiple, the sum of the currents is the same as it was in one wire. But the voltage won't be altered in "cross roads".

    3) You need to find the Req for the circuit and then you can use the equation for power.
  4. Feb 21, 2009 #3
    your right there is an equation for resistivity, that includes lenght and cross sectional area.
    but i dont know what the restistivy of my matiral is.

    2) for the series circuit would the current through each resistor be 2?
    B)i dont get you

    3)thanks for nothing on that one, lol you just read the question too me.
  5. Feb 21, 2009 #4
    That's what your trying to figure out. The equation should be [tex]R= \rho \frac{l}{A}[/tex] you know everything else but [tex]\rho[/tex].


    Then you should read about Kirchoff's law. But to get you going, the voltage through each resistor is the same, current isn't.

    Your welcome! :tongue2: http://en.wikipedia.org/wiki/Resistor#Series_and_parallel_resistors Check that out. You should see those type of connections in the circuit, which you then can use to determine Req for the circuit.
  6. Feb 21, 2009 #5
    lol thanks man, didnt mean too sound like a loser.
  7. Feb 21, 2009 #6
    i still dont see how my curent through each resistor is wrong for the parallel cirucit.
  8. Feb 21, 2009 #7
    Try drawing the circuit out with the 4 resistances in parallel connected to the voltage source. You should be able to see that the voltage across each resistor is the same as the voltage source. So if you know the voltage across each resistor, and the resistance of each resistor, you should be able to solve for the current through each resistor.
  9. Feb 21, 2009 #8
    Start collapsing the circuit from the outside, moving towards the voltage source. Use parallel combination where you need it, and use series combination where you need it. It may help to redraw the circuit each time you make a combination, in order to easily see subsequent combinations. Once the circuit is down to just a voltage source and an equivalent resistor, it should be easy to find the power. There are multiple ways to find it.
  10. Feb 21, 2009 #9
    isnt the formula i=V/R1 , if thats true then for resisotr one it would be 24/1
  11. Feb 21, 2009 #10
    Yep, that's one branch current, just get the other 3 for that parallel circuit and you should be set for that part. After you've calculated those currents, you will need to re-figure the equivalent parallel resistance REQ, it is not 1/3 ohms but it's close.

    Once you get the correct REQ, calculate the current through it. You will notice that the current through the REQ is the sum of the individual branch currents. In other words: IREQ = IR1 + IR2 + IR3 + IR4
    It might be a good way to confirm your answer.
  12. Feb 22, 2009 #11
    is it .5?

    for number 3 is the equivalent resistance 6?
  13. Feb 22, 2009 #12
    Yep, both are right, good job :biggrin:
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