10.2 Resistors in series and parallel  (Page 6/11)

In this chapter, we introduced the equivalent resistance of resistors connect in series and resistors connected in parallel. You may recall that in Capacitance , we introduced the equivalent capacitance of capacitors connected in series and parallel. Circuits often contain both capacitors and resistors. [link] summarizes the equations used for the equivalent resistance and equivalent capacitance for series and parallel connections.

Combinations of series and parallel

More complex connections of resistors are often just combinations of series and parallel connections. Such combinations are common, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel.

Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in [link] . Various parts can be identified as either series or parallel connections, reduced to their equivalent resistances, and then further reduced until a single equivalent resistance is left. The process is more time consuming than difficult. Here, we note the equivalent resistance as $R_{\text{eq}}.$

Notice that resistors $R_3$ and $R_4$ are in series. They can be combined into a single equivalent resistance. One method of keeping track of the process is to include the resistors as subscripts. Here the equivalent resistance of $R_3$ and $R_4$ is

The circuit now reduces to three resistors, shown in [link] (c). Redrawing, we now see that resistors $R_2$ and $R_{34}$ constitute a parallel circuit. Those two resistors can be reduced to an equivalent resistance:

This step of the process reduces the circuit to two resistors, shown in in [link] (d). Here, the circuit reduces to two resistors, which in this case are in series. These two resistors can be reduced to an equivalent resistance, which is the equivalent resistance of the circuit:

The main goal of this circuit analysis is reached, and the circuit is now reduced to a single resistor and single voltage source.

Now we can analyze the circuit. The current provided by the voltage source is $I=\frac{V}{R_{\text{eq}}}=\frac{24\text{V}}{12\text{Ω}}=2\text{A}.$ This current runs through resistor $R_1$ and is designated as $I_1.$ The potential drop across $R_1$ can be found using Ohm’s law: