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

1. Jan 13, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

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

http://img215.imageshack.us/img215/3074/chapter2problem38aw0.jpg [Broken]

2. Relevant equations

$$v\,=\,i\,R$$

Parallel and series resistor equations.

3. 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 [Broken]

$$v\,=\,i\,R$$

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

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

So, $R_{eq}\,=\,10.3\Omega$ and $i_0\,=\,8\,A$?

Last edited by a moderator: May 2, 2017
2. Jan 13, 2007

### 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.

3. Jan 13, 2007

### VinnyCee

Whoops, maybe $R_{eq}\,=\,7.5\Omega$?

http://img90.imageshack.us/img90/1933/chapter2problem38part3jr5.jpg [Broken]

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 [Broken]

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 [Broken]

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

http://img20.imageshack.us/img20/8697/chapter2problem38part6bu7.jpg [Broken]

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 [Broken]

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 [Broken]

So how do I solve for $i_0$ if that is the right $R_{eq}$?

Last edited by a moderator: May 2, 2017
4. Jan 13, 2007

### 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.

5. Jan 13, 2007

### VinnyCee

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

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

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

Last edited: Jan 13, 2007
6. Jan 13, 2007

### teknodude

yea that's correct