CIRCUIT ANALYSIS: Find the equivalent resistance seen by the source

1. Jan 12, 2007

VinnyCee

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.

2. Relevant equations

For resistors in series: $$R_{eq}\,=\,R_1\,+\,R_2\,+\,\cdots\,+\,R_n$$

For resistors in parallel: $$R_{eq}\,=\,\frac{1}{R_1}\,+\,\frac{1}{R_2}\,+\,\cdots\,+\,\frac{1}{R_n}$$

Also for resistors in parallel: $$R_{eq}\,=\,\frac{R_1\,R_2}{R_1\,+\,R_2}$$

3. The attempt at a solution

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

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

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

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

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

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

$$i\,=\,\frac{v}{R}\,=\,\frac{12\,V}{40\Omega}\,=\,0.3\,A$$

The I use the p = vi equation.

$$p\,=\,v\,i\,=\,\left(12\,V\right)\,\left(0.3\,A\right)\,=\,3.6\,W$$

Right?

Last edited: Jan 12, 2007
2. Jan 13, 2007

doodle

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