a) A network of five equal resistors R is connected to a battery \xi as shown in Fig. 19–60. Determine the current I that flows out of the battery. b) Use the value determined for I to find the single resistor R_{eq} that is equivalent to the five-resistor network.

Giancoli 7th Edition, Chapter 19, Problem 33

Problem 33

A network of five equal resistors $R$ is connected to a battery $\xi$ as shown in Fig. 19–60. Determine the current $I$ that flows out of the battery.
Use the value determined for $I$ to find the single resistor $R_{eq}$ that is equivalent to the five-resistor network.

Figure 19-60.

$I = \dfrac{\xi}{R}$
$R$

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COMMENTS

By livanessa98 on Tue, 7/5/2016 - 4:52 PM

Two quick questions:
1) Why is the answer for part b different from what you would find by doing Req calculations based on the rules for parallel and in-series resistors? (I got 5R/8 that way).
2) By considering all three resistors that are arranged in series as having the same current by definition, you could solve for I with 3 equations and 3 unknowns. Would this be a correct approach?

By Mr. Dychko on Wed, 7/6/2016 - 7:09 PM

Hi livanessa98, good questions.

1) The series/parallel formulas don't apply here. Looking at the circuit diagram, the junction with $I_2$ going in and $I_3$ and $I_4$ coming out makes all the difference. None of the resistors in this picture are in series. Surprising, I know! When resistors are in series it means that all the current going through one resistor must also go through the next, but with these junctions interfering, some of the current through one is being diverted among multiple resistors downstream, and for this reason they're not in series. None of the resistors are in parallel either because of these pesky junctions, which makes it a clever circuit diagram.
2) See (1) :)

Best wishes,
Mr. Dychko

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