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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.
[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
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