Problem 2.16   Source:   Institute of Physics - Ufrgs - Basic Eletronic discipline - List of exercises - Prof. Dr. Francis

In the circuit shown in Figure 02-16.1 find the equivalent resistance seen from the terminals a-b when:

a) the c -d terminals are open.

b) the c -d terminals are closed.

Solution of the Problem 2.16

Item a

When the c-d points are open, the R₄ and R₆ as well as R₅ and R₇, are in series. Therefore, we can add their values. So R₄ + R₆ = 10 ohms and R₅ + R₇ = 15 ohms . The circuit in the Figure 02-16.2 shows the modifications with their respective values.

It is now easy to see that we get three resistors in parallel. So using the equation for the calculation of resistances in parallel we find R_p = 3.75 ohms as shown in the circuit in Figure 02-16.3.

And finally we were able to reduce the original circuit to a series of three resistors, making our calculations easier. So to find the value of R_eq, just add the value of the resistors that make up the circuit, or:

Item b

We must now calculate the equivalent resistance of the circuit, considering that points c - d are shorted. In this case, the resistances R₄ and R₅ are connected in parallel as well as R₆ and R₇. Thus, performing the calculations we find:

So these two parallels are in series, as we can see in the Figure 02-16.4.

Note that the two resistors in series, R₄₅ and R₆₇, total 5 ohms , which in turn are parallel to R₂ . Calculating the value of this parallel, we have:

As we see in theFigure 02-16.5, we get three resistors in series. So, to find the equivalent resistance, we add its values and we found: