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CIRCUIT ANALYSIS: Find the equivalent resistance seen by the source

  • Engineering
  • Thread starter VinnyCee
  • Start date
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1. Homework Statement

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. Homework 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.

[tex]i\,=\,\frac{v}{R}\,=\,\frac{12\,V}{40\Omega}\,=\,0.3\,A[/tex]

The I use the p = vi equation.

[tex]p\,=\,v\,i\,=\,\left(12\,V\right)\,\left(0.3\,A\right)\,=\,3.6\,W[/tex]

Right?
 
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Answers and Replies

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Or equivalently, P = V^2/R which gives the same answer.
 

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