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anonymous

  • one year ago

A 12.0 V battery is hooked up with three resistors (R1, R2, and R3) in parallel with resistances of 2.5 Ω, 7.0 Ω, and 15.0 Ω, respectively. Part 1: Draw a labeled circuit diagram for the circuit described using correct symbols. Part 2: Calculate the equivalent resistance. Part 3: Calculate the current passing through each resistor in the circuit. ***How do I do this?

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  1. Michele_Laino
    • one year ago
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    the requested circuit is: |dw:1434903805092:dw|

  2. anonymous
    • one year ago
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    so that is the drawing?

  3. Michele_Laino
    • one year ago
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    |dw:1434903890518:dw|

  4. Michele_Laino
    • one year ago
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    yes! I think so!

  5. anonymous
    • one year ago
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    ok yay!! so that is it for part 1?

  6. Michele_Laino
    • one year ago
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    yes!

  7. anonymous
    • one year ago
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    ookie yay!! and so now we go to part 2? how do we do that? :/

  8. Michele_Laino
    • one year ago
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    part #2 the equivalent resistance is: \[\Large \begin{gathered} {R_{EQUIVALENT}} = \frac{1}{{\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}}} = \hfill \\ \hfill \\ = \frac{1}{{\frac{1}{{2.5}} + \frac{1}{7} + \frac{1}{{15}}}} = ...Ohm \hfill \\ \end{gathered} \]

  9. anonymous
    • one year ago
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    ok! so we get 1.640625?

  10. Michele_Laino
    • one year ago
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    that's right!

  11. anonymous
    • one year ago
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    yay so that is our equivalent resistance? :O

  12. Michele_Laino
    • one year ago
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    it is 1.64 ohms

  13. anonymous
    • one year ago
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    ooh okie yay!! so onto part 3!

  14. Michele_Laino
    • one year ago
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    the requested currents are: \[\Large \begin{gathered} {I_1} = \frac{{12}}{{2.5}} = ...Amps \hfill \\ \hfill \\ {I_2} = \frac{{12}}{7} = ...Amps \hfill \\ \hfill \\ {I_3} = \frac{{12}}{{15}} = ...Amps \hfill \\ \end{gathered} \]

  15. Michele_Laino
    • one year ago
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    |dw:1434904298434:dw|

  16. anonymous
    • one year ago
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    ok! we get these values? 4.8 1.71 0.8 ?

  17. Michele_Laino
    • one year ago
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    correct!

  18. anonymous
    • one year ago
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    yay!! and so we are finished with this problem? :O

  19. Michele_Laino
    • one year ago
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    yes!

  20. anonymous
    • one year ago
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    ooh yay!! thank you!!:) onto the next!!

  21. Michele_Laino
    • one year ago
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    :) ok!

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