9.1 Resistors in series and parallel  (Page 5/17)

Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in [link] . Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult.

The simplest combination of series and parallel resistance, shown in [link] , is also the most instructive, since it is found in many applications. For example, $R_1$ could be the resistance of wires from a car battery to its electrical devices, which are in parallel. $R_2$ and $R_3$ could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates.

Calculating resistance, $\text{IR}$ Drop, current, and power dissipation: combining series and parallel circuits

[link] shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider $R_1$ to be the resistance of wires leading to $R_2$ and $R_3$ . (a) Find the total resistance. (b) What is the $\text{IR}$ drop in $R_1$ ? (c) Find the current $I_2$ through $R_2$ . (d) What power is dissipated by $R_2$ ?

Strategy and Solution for (a)

To find the total resistance, we note that $R_2$ and $R_3$ are in parallel and their combination $R_{\text{p}}$ is in series with $R_1$ . Thus the total (equivalent) resistance of this combination is

First, we find $R_{\text{p}}$ using the equation for resistors in parallel and entering known values:

Inverting gives

So the total resistance is

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values ( $\mathrm{20.0\; \Omega }$ and $\mathrm{0.804\; \Omega }$ , respectively) found for the same resistors in the two previous examples.

Strategy and Solution for (b)

To find the $\text{IR}$ drop in $R_1$ , we note that the full current $I$ flows through $R_1$ . Thus its $\text{IR}$ drop is

We must find $I$ before we can calculate $V_1$ . The total current $I$ is found using Ohm’s law for the circuit. That is,

Entering this into the expression above, we get

Discussion for (b)

The voltage applied to $R_2$ and $R_3$ is less than the total voltage by an amount $V_1$ . When wire resistance is large, it can significantly affect the operation of the devices represented by $R_2$ and $R_3$ .

Strategy and Solution for (c)

To find the current through $R_2$ , we must first find the voltage applied to it. We call this voltage $V_{\text{p}}$ , because it is applied to a parallel combination of resistors. The voltage applied to both $R_2$ and $R_3$ is reduced by the amount $V_1$ , and so it is