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

[VinnyCee]

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 $i_0\,=\,8\,A$?

 

[doodle]

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.

 

[VinnyCee]

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 $R_{eq}$?

 

[BobG]

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.

 

[VinnyCee]

$$V\,=\,i\,R$$

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

So, $i_0\,=\,3.2\,A$?

 

[teknodude]

yea that's correct