21.1 Resistors in series and parallel  (Page 6/17)

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${V}_{\text{p}}=V-{V}_{1}=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}-2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{V}=9\text{.}\text{65}\phantom{\rule{0.25em}{0ex}}\text{V}.$

Now the current ${I}_{2}$ through resistance ${R}_{2}$ is found using Ohm’s law:

${I}_{2}=\frac{{V}_{\text{p}}}{{R}_{2}}=\frac{9\text{.}\text{65 V}}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }=1\text{.}\text{61}\phantom{\rule{0.25em}{0ex}}\text{A}.$

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

${P}_{2}=\left({I}_{2}{\right)}^{2}{R}_{2}=\left(1\text{.}\text{61}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=\text{15}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{W}.$

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.

Practical implications

One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the $\text{IR}$ drop in the wires can also be significant.

For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in [link] . The device represented by ${R}_{3}$ has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger $\text{IR}$ drop in the wires represented by ${R}_{1}$ , reducing the voltage across the light bulb (which is ${R}_{2}$ ), which then dims noticeably.

Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel.

No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff’s rules, to be introduced in Kirchhoff’s Rules , will allow you to analyze the circuit.

Problem-solving strategies for series and parallel resistors

1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they are labeled in your circuit diagram.
2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them.
4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series or parallel connections, as done in this module and the examples. Special note: When finding $R{}_{\text{p}}\text{}$ , the reciprocal must be taken with care.
5. Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas total parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on.

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