General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 17, Problem 70E
To determine

The equivalent resistance of the network shown in Figure 17.30.

Expert Solution & Answer
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Answer to Problem 70E

The equivalent resistance of the network shown in Figure 17.30 is 4r_.

Explanation of Solution

The resistor network is shown in Figure 17.30.

The resistances are labelled as R1, R2, R3, R4, R5, R6, R7, R8, and R9 as shown in Figure 1.

General Physics, 2nd Edition, Chapter 17, Problem 70E

The resistors R7, R8, and R9 are in series, so that their total resistance is given by,

  R7,8,9=R7+R8+R9        (I)

The resistor R6 and the combination R7,8,9 are in parallel with total resistance,

  R6,7,8,9=R6R7,8,9R6+R7,8,9        (II)

The resistors R4, R5, and R6,7,8,9 are in series. So their net resistance is given by,

  R4,5,6,7,8,9=R4+R5+R6,7,8,9        (III)

The resistor R3 and R4,5,6,7,8,9 are in parallel, with total resistance,

  R3,4,5,6,7,8,9=R3R4,5,6,7,8,9R3+R4,5,6,7,8,9        (IV)

The resistors R1, R2, and R3,4,5,6,7,8,9 are in series. Thus, the equivalent resistance of the network is given by,

  Requivalent=R1+R2+R3,4,5,6,7,8,9        (V)

Conclusion:

Substitute 2r for R7, R8, and R9 in equation (I) to find R7,8,9.

  R7,8,9=2r+2r+2r=6r

Substitute 3r for R6, and 6r for R7,8,9 in equation (II) to find R6,7,8,9.

  R6,7,8,9=(3r)(6r)3r+6r=2r

Substitute 2r for R4, R5, and R6,7,8,9 in equation (III) to find R4,5,6,7,8,9.

  R4,5,6,7,8,9=2r+2r+2r=6r

Substitute 3r for R3,and 6r for R4,5,6,7,8,9 in equation (IV) to find R3,4,5,6,7,8,9.

  R3,4,5,6,7,8,9=(3r)(6r)3r+6r=2r

Substitute r for R1 and R2, and 2r for R3,4,5,6,7,8,9 in equation (V) to find Requivalent.

  Requivalent=r+r+2r=4r

Therefore, the equivalent resistance of the network shown in Figure 17.30 is 4r_.

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Chapter 17 Solutions

General Physics, 2nd Edition

Ch. 17 - Prob. 11RQCh. 17 - Prob. 12RQCh. 17 - Prob. 1ECh. 17 - Prob. 2ECh. 17 - Prob. 3ECh. 17 - Prob. 4ECh. 17 - Prob. 5ECh. 17 - Prob. 6ECh. 17 - Prob. 7ECh. 17 - Prob. 8ECh. 17 - Prob. 9ECh. 17 - Prob. 10ECh. 17 - Prob. 11ECh. 17 - Prob. 12ECh. 17 - Prob. 13ECh. 17 - Prob. 14ECh. 17 - Prob. 15ECh. 17 - Prob. 16ECh. 17 - Prob. 17ECh. 17 - Prob. 18ECh. 17 - Prob. 19ECh. 17 - Prob. 20ECh. 17 - Prob. 21ECh. 17 - Prob. 22ECh. 17 - Prob. 23ECh. 17 - Prob. 24ECh. 17 - Prob. 25ECh. 17 - Prob. 26ECh. 17 - Prob. 27ECh. 17 - Prob. 28ECh. 17 - Prob. 29ECh. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Prob. 33ECh. 17 - Prob. 34ECh. 17 - Prob. 35ECh. 17 - Prob. 36ECh. 17 - Prob. 37ECh. 17 - Prob. 38ECh. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Prob. 41ECh. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Prob. 44ECh. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - Prob. 50ECh. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Prob. 53ECh. 17 - Prob. 54ECh. 17 - Prob. 55ECh. 17 - Prob. 56ECh. 17 - Prob. 57ECh. 17 - Prob. 58ECh. 17 - Prob. 59ECh. 17 - Prob. 60ECh. 17 - Prob. 61ECh. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - Prob. 67ECh. 17 - Prob. 68ECh. 17 - Prob. 69ECh. 17 - Prob. 70ECh. 17 - Prob. 71ECh. 17 - Prob. 72ECh. 17 - Prob. 73ECh. 17 - Prob. 74ECh. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - Prob. 80ECh. 17 - Prob. 81ECh. 17 - Prob. 82ECh. 17 - Prob. 83ECh. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Prob. 87ECh. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - Prob. 92E
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