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CIRCUIT ANALYSIS: Find the equivalent resistance seen by the source

  1. Jan 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit below. Find the overall dissipated power.

    http://img300.imageshack.us/img300/6132/chapter2problem34au4.jpg [Broken]

    2. Relevant equations

    For resistors in series: [tex]R_{eq}\,=\,R_1\,+\,R_2\,+\,\cdots\,+\,R_n[/tex]

    For resistors in parallel: [tex]R_{eq}\,=\,\frac{1}{R_1}\,+\,\frac{1}{R_2}\,+\,\cdots\,+\,\frac{1}{R_n}[/tex]

    Also for resistors in parallel: [tex]R_{eq}\,=\,\frac{R_1\,R_2}{R_1\,+\,R_2}[/tex]

    3. The attempt at a solution

    Adding the right three resistors that are in series to get an equivalent one that is 40 Ohms.

    http://img205.imageshack.us/img205/1186/chapter2problem34part2dk8.jpg [Broken]

    Using the formula above for parallel resistors: [tex]R_{eq}\,=\,\frac{(40\Omega)\,(40\Omega)}{(40\Omega)\,+\,(40\Omega)}\,=\, 20\Omega[/tex]

    http://img391.imageshack.us/img391/7092/chapter2problem34part3ur9.jpg [Broken]

    Again, combining the resistors on the right that are in series.

    http://img205.imageshack.us/img205/5549/chapter2problem34part4jn1.jpg [Broken]

    Using the parallel formula again: [tex]R_{eq}\,=\,\frac{(40\Omega)\,(40\Omega)}{(40\Omega)\,+\,(40\Omega)}\,=\, 20\Omega[/tex]

    http://img397.imageshack.us/img397/9507/chapter2problem34part5su9.jpg [Broken]

    Finally, adding the last two resistors in series, I get an [itex]R_{eq}[/itex] of [itex]40\Omega[/itex]. Does this seem correct?

    http://img300.imageshack.us/img300/2751/chapter2problem34part6ma8.jpg [Broken]

    Then, to find the Power, I get the current first.


    The I use the p = vi equation.


    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 13, 2007 #2
    Or equivalently, P = V^2/R which gives the same answer.
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