Engineering CIRCUIT ANALYSIS: 7resistors & 1 I.V.S. - Find equivalent resistance and current

AI Thread Summary
The discussion focuses on calculating the equivalent resistance (R_eq) and current (i_0) in a circuit with seven resistors and one independent voltage source. Initial calculations suggested R_eq as 10.3Ω and i_0 as 8A, but further analysis indicated that R_eq should be 7.5Ω. After combining resistors in parallel and series, the correct equivalent resistance was determined to be 12.5Ω. Using this value, the current i_0 was recalculated to be 3.2A. The final consensus confirmed that i_0 equals 3.2A based on the reduced circuit analysis.
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Homework Statement



Find R_{eq} and i_0 in the circuit below.

http://img215.imageshack.us/img215/3074/chapter2problem38aw0.jpg

Homework Equations



v\,=\,i\,R

Parallel and series resistor equations.

The Attempt at a Solution



After using the resistor equations to get down to R_{eq}, I get the diagram below.

http://img218.imageshack.us/img218/3425/chapter2problem38part2xf8.jpg

v\,=\,i\,R

40\,V\,=\,i_0\,\left(5\Omega\right)

i_0\,=\,\frac{40\,V}{5\Omega}\,=\,8\,A

So, R_{eq}\,=\,10.3\Omega and i_0\,=\,8\,A?
 
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I don't agree that 10.3ohm is the result of the combination of the other 6 resistors.

In the second figure, surely you cannot write i0 = 40/5 without taking into account the other (10.3ohm) resistor.
 
Whoops, maybe R_{eq}\,=\,7.5\Omega?

http://img90.imageshack.us/img90/1933/chapter2problem38part3jr5.jpg

Combining the two resistors in parallel that are circled in green.

\frac{(12\Omega)\,(6\Omega)}{(12\Omega)\,+\,(6\Omega)}\,=\,\frac{72}{18}\,\Omega\,=\,4\Omega

http://img111.imageshack.us/img111/9434/chapter2problem38part4um9.jpg

Now combining the two resistors circled in green that are in parallel.

\frac{(80\Omega)\,(20\Omega)}{(80\Omega)\,+\,(20\Omega)}\,=\,\frac{1600}{100}\,\Omega\,=\,16\Omega

http://img441.imageshack.us/img441/9001/chapter2problem38part5rb8.jpg

Combine the two resistors circled in green that are in series.

http://img20.imageshack.us/img20/8697/chapter2problem38part6bu7.jpg

Combine the two resistors circled in green that are in parallel.

\frac{(60\Omega)\,(20\Omega)}{(60\Omega)\,+\,(20\Omega)}\,=\,\frac{1200}{80}\,\Omega\,=\,15\Omega

http://img294.imageshack.us/img294/7270/chapter2problem38part7ac4.jpg

Combine the last two resistors circled in green that are in parallel.

\frac{(15\Omega)\,(15\Omega)}{(15\Omega)\,+\,(15\Omega)}\,=\,\frac{225}{30}\,\Omega\,=\,7.5\Omega

http://img363.imageshack.us/img363/1525/chapter2problem38part8yh6.jpg

So how do I solve for i_0 if that is the right R_{eq}?
 
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In your reduced circuit, the current should be the same through the whole circuit, right? Which makes sense, since the sum of the currents through all of those branches has to equal the current going through the 5 ohm resisitor.

You've reduced your circuit to a series circuit. Divide the voltage by the sum of your resistance.
 
V\,=\,i\,R

i_0\,=\,\frac{V}{R}\,=\,\frac{40\,V}{12.5\Omega}\,=\,3.2\,A

So, i_0\,=\,3.2\,A?
 
Last edited:
yea that's correct
 

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