Combinations of Series and Parallel
Combinations of Series and Parallel
More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, 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 this figure. 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.
This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached.
The simplest combination of series and parallel resistance, shown in this figure, 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.
Example: Calculating Resistance, \(\text{IR}\) Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits
This figure 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 \({R}_{2}\)?
These three resistors are connected to a voltage source so that \({R}_{2}\) and \({R}_{3}\) are in parallel with one another and that combination is in series with \({R}_{1}\).
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
\({R}_{\text{tot}}={R}_{1}+{R}_{\text{p}}.\)
First, we find \({R}_{\text{p}}\) using the equation for resistors in parallel and entering known values:
\(\cfrac{1}{{R}_{\text{p}}}=\cfrac{1}{{R}_{2}}+\cfrac{1}{{R}_{3}}=\cfrac{1}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }+\cfrac{1}{\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega }=\cfrac{0\text{.}\text{2436}}{\Omega }.\)
Inverting gives
\({R}_{\text{p}}=\cfrac{1}{0\text{.}\text{2436}}\Omega =4\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega .\)
So the total resistance is
\({R}_{\text{tot}}={R}_{1}+{R}_{\text{p}}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega +4\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega =5\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega .\)
Discussion for (a)
The total resistance of this combination is intermediate between the pure series and pure parallel values (\(20.0 \Omega \) and \(0.804 \Omega \), 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
\({V}_{1}={\text{IR}}_{1}.\)
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,
\(I=\cfrac{V}{{R}_{\text{tot}}}=\cfrac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{5\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega }=2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{A}.\)
Entering this into the expression above, we get
\({V}_{1}={\text{IR}}_{1}=(2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{A})(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega )=2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{V}.\)
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
\({V}_{\text{p}}=V-{V}_{1}=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}-2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{V}=9\text{.}\text{65}\phantom{\rule{0.25em}{0ex}}\text{V}.\)
Now the current \({I}_{2}\) through resistance \({R}_{2}\) is found using Ohm’s law:
\({I}_{2}=\cfrac{{V}_{\text{p}}}{{R}_{2}}=\cfrac{9\text{.}\text{65 V}}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }=1\text{.}\text{61}\phantom{\rule{0.25em}{0ex}}\text{A}.\)
Discussion for (c)
The current is less than the 2.00 A that flowed through \({R}_{2}\) when it was connected in parallel to the battery in the previous parallel circuit example.
Strategy and Solution for (d)
The power dissipated by \({R}_{2}\) is given by
\({P}_{2}=({I}_{2}{)}^{2}{R}_{2}=(1\text{.}\text{61}\phantom{\rule{0.25em}{0ex}}\text{A}{)}^{2}(6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega )=\text{15}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{W}.\)
Discussion for (d)
The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.
This lesson is part of:
Circuits and DC Instruments