Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 19, Problem 11P
To determine

The four different ways for connection of resistors and equivalent resistance in each case.

Expert Solution & Answer
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Explanation of Solution

Given info:

The resistance of the resistor is, R1=R2=R3=240Ω .

The three resistors may be connected as all three in series, all three in parallel, one in series and other two in parallel and one in parallel with series combination of the remaining two resistors.

Formula Used:

The expression to calculate the equivalent resistance of series connection of resistors is,Rs=R1+R2+R3

The expression to calculate the equivalent resistance of series connection of resistors is,Rp=R1R2R3R1R2+R2R3+R3R1

The expression to calculate the equivalent resistance of one resistance in series of other two parallel resistances is,Re12=R1+R2R3R2+R3

The expression to calculate the equivalent resistance of one resistance in parallel of other two series resistances is,Rp1=R1(R2+R3)R1+R2+R3

Calculation:

Substitute all the values in the above expression.

  Rs=240Ω+240Ω+240Ω=720Ω

Substitute all the values in the above expression.

  Rp=(240Ω)(240Ω)(240Ω)(240Ω)(240Ω)+(240Ω)(240Ω)+(240Ω)(240Ω)=80Ω

Substitute all the values in the above expression.

  Re1=240Ω+(240Ω)(240Ω)240Ω+240Ω=360Ω

Substitute all the values in the above expression.

  Rp1=(240Ω)(240Ω+240Ω)240Ω+240Ω+240Ω=160Ω

Conclusion:

Therefore, the four ways of connections of resistors and equivalent resistance of each way are following:

    Sr.No.Way of connectionEquivalent resistance
    1All three in series connection720Ω
    2All three in parallel connection80Ω
    3One in series with other two in parallel connection360Ω
    4One in parallel with other two in series connection160Ω

Chapter 19 Solutions

Physics: Principles with Applications

Ch. 19 - Prob. 11QCh. 19 - Prob. 12QCh. 19 - Prob. 13QCh. 19 - Prob. 14QCh. 19 - Prob. 15QCh. 19 - Prob. 16QCh. 19 - Given the circuit shown in Fig. 19-38, use the...Ch. 19 - Prob. 18QCh. 19 - Prob. 19QCh. 19 - 19. What is the main difference between an analog...Ch. 19 - What would happen if you mistakenly used an...Ch. 19 - Prob. 22QCh. 19 - Prob. 23QCh. 19 - Prob. 24QCh. 19 - Prob. 1PCh. 19 - Prob. 2PCh. 19 - Prob. 3PCh. 19 - Prob. 4PCh. 19 - Prob. 5PCh. 19 - Prob. 6PCh. 19 - Prob. 7PCh. 19 - Prob. 8PCh. 19 - Prob. 9PCh. 19 - Prob. 10PCh. 19 - Prob. 11PCh. 19 - Prob. 12PCh. 19 - Prob. 13PCh. 19 - Prob. 14PCh. 19 - Prob. 15PCh. 19 - Prob. 16PCh. 19 - Prob. 17PCh. 19 - Prob. 18PCh. 19 - Prob. 19PCh. 19 - Prob. 20PCh. 19 - Prob. 21PCh. 19 - Prob. 22PCh. 19 - Prob. 23PCh. 19 - Prob. 24PCh. 19 - Prob. 25PCh. 19 - Prob. 26PCh. 19 - Prob. 27PCh. 19 - Prob. 28PCh. 19 - Prob. 29PCh. 19 - Prob. 30PCh. 19 - Prob. 31PCh. 19 - Prob. 32PCh. 19 - Prob. 33PCh. 19 - Prob. 34PCh. 19 - Prob. 35PCh. 19 - Prob. 36PCh. 19 - Prob. 37PCh. 19 - Prob. 38PCh. 19 - Prob. 39PCh. 19 - Prob. 40PCh. 19 - Prob. 41PCh. 19 - Prob. 42PCh. 19 - Prob. 43PCh. 19 - Prob. 44PCh. 19 - Prob. 45PCh. 19 - Prob. 46PCh. 19 - Prob. 47PCh. 19 - Prob. 48PCh. 19 - Prob. 49PCh. 19 - Prob. 50PCh. 19 - Prob. 51PCh. 19 - Prob. 52PCh. 19 - Prob. 53PCh. 19 - Prob. 54PCh. 19 - Prob. 55PCh. 19 - Prob. 56PCh. 19 - Prob. 57PCh. 19 - Prob. 58PCh. 19 - Prob. 59PCh. 19 - Prob. 60PCh. 19 - Prob. 61PCh. 19 - Prob. 62PCh. 19 - Prob. 63PCh. 19 - Prob. 64GPCh. 19 - Prob. 65GPCh. 19 - Prob. 66GPCh. 19 - Prob. 67GPCh. 19 - Prob. 68GPCh. 19 - Prob. 69GPCh. 19 - Prob. 70GPCh. 19 - Prob. 71GPCh. 19 - Prob. 72GPCh. 19 - Prob. 73GPCh. 19 - Prob. 74GPCh. 19 - Prob. 75GPCh. 19 - Prob. 76GPCh. 19 - Prob. 77GPCh. 19 - Prob. 78GPCh. 19 - Prob. 79GPCh. 19 - Prob. 80GPCh. 19 - Prob. 81GPCh. 19 - Prob. 82GPCh. 19 - Prob. 83GPCh. 19 - Prob. 84GPCh. 19 - Prob. 85GPCh. 19 - Prob. 86GPCh. 19 - Prob. 87GP

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