Problem 2.8   Source:   Problem elaborated by the author of the site.

Calculate the equivalent resistance between the points a-b of the circuit shown in Figure 02-08.1.

Solution of the Problem 2.8

In the circuit above we number the nodes to better understand the solution. Between the points 1 - 2 we have two resistors in parallel. One of 12 ohms and one of 36 ohms . Thus the equivalent resistance due to the parallel of these two resistors, which we will call R₁₂, is equal to R₁₂ = 12 x 36 / (12 + 36) = 9 ohms . Note that by replacing the two resistors in parallel with R₁₂ , we have the same circuit as the previous problem. Thus, the solution sequence follows the same steps. So R₁₂ is in series with the other resistor, which we will call R₂₃ = 7 ohms, totaling R '₁₃ = 7 + 9 = 16 ohms. And this resistor is parallel to the other resistor of 16 ohms that is between the points 1 - 3. So we have an equivalent resistance between the points 1 - 3, which we will call R₁₃ , equal to:

Thus all resistors that lie between points 1-3 can be replaced by a single resistor value of R₁₃ = 8 ohms. And this resistor is in series with that found between the points 3-4, which we will call R₃₄. Soon, we can sum them up and get the value of R₁₃₄ = 20 ohms . It is easy to see that R₁₃₄ is parallel to the resistor that is between points 1 - 4, which we will call R '₁₄. Calculating the parallel of the two we find the value of:

And finally we have three resistors in series, resulting in an equivalent resistance between points a-b of:

This means that the entire circuit shown in the figure can be replaced by a single resistor of value equal to 48 ohms.