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Analyzing circuits with resistor/battery

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    use circuit shown:

    if my 'drawing' gets reformatted incorrectly, this is what it 'looks' like in words.
    a 25 volt battery connected to a 16ohm at the top which connects at a junction to a 8ohm that goes down and a 1 ohm that goes to the right. the 1 ohm forms top right corner with a 3 ohm resistor pointing down to the bottom left corner. the bottom left corner connects to a4 ohm resistor that connects with 8 ohm resistor coming down from the 16 ohm - 1 ohm wire. the 4 ohm is connected at a junction to a 5 ohm resistor which connects to the bottom of the battery.

    the battery, 8ohm, and 3 ohm are parallel. the 16ohm and 1 ohm wire is parallel to the 5ohm-4ohm wire. both are perpendicular to the battery, 8ohm and 3 ohm resistor.

    btw, for future reference, what is the best way to put up diagrams like circuit so that it won't get messed up during posting?

    _____16ohm__________1ohm___
    | | |
    --- | |
    25volts 8 3
    - o o
    | h h
    | m m
    |____5ohm__|________4ohm___|


    a) what is the potential difference accross the 4 ohm resistor in the circuit?
    b) what is the current in the 8 ohm resistor?
    c) at what rate is the energy dissipated in the 16 ohm resistor and what fractions of all the energy dissipated is this?


    2. Relevant equations

    series resistors R = R_1 + R_2 + R_3 ...

    parallel resistors 1/R = 1/R_1 + 1/R_2 + 1/R_3 ....

    V = IR where V is potential, I is current, R is resistance

    current I = epsilon/(R_1 + R_2) ---> not sure what epsilon stands for? R is resistor


    3. The attempt at a solution

    i want to know the best way to handle the circuit, i am not sure what to make into series or parallel resistors to simplify the circuit.

    especially how to handle the 8 and 3 ohm resistors in the center

    help appreciated
     

    Attached Files:

    Last edited: Jun 11, 2008
  2. jcsd
  3. Jun 11, 2008 #2

    Tom Mattson

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    There are 3 resistors that are obviously in series (that is, the same current flows through all 3 of them). Can you spot them?
     
  4. Jun 11, 2008 #3
    the 16 ohm, 1 ohm and 3 ohm resistors?

    if i made those three in series then how would the 8 ohm and 4 ohm be involved when i redraw the circuit?

    attach a screen shot because my 'drawing' got messed up..if it helps
     
    Last edited: Jun 11, 2008
  5. Jun 11, 2008 #4

    Tom Mattson

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    No, the 16 ohm resistor is clearly not in series with the others. The current flowing through that resistor gets split off into the 8 ohm resistor the branch containing the 1, 3, and 4 ohm resistors.

    Try again?
     
  6. Jun 11, 2008 #5
    the 1, 3, 4 ohm resistors? and then the 16, 8, 5 ohm resistors, then both series joined in parallel?
     
  7. Jun 17, 2008 #6

    Tom Mattson

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    Hi, sorry for the delay. I've been busy.

    Yes, the 1,3, and 4 ohm resistors are in series. But the rest are not in parallel. If you reduce those 3 resistors to a single resistor, you should be able to see that it is in parallel with the 8 ohm resistor.

    Can you take it from there?
     
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