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 71E

(a)

To determine

The current in the EMF in the circuit shown in Figure 17.31.

(a)

Expert Solution
Check Mark

Answer to Problem 71E

The current in the EMF in the circuit shown in Figure 17.31 is ε/r_.

Explanation of Solution

The resistor network is shown in Figure 17.31.

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

General Physics, 2nd Edition, Chapter 17, Problem 71E

The resistors R1, R2, and R3 are in parallel, so that their total resistance can be found out as,

  1R1,2,3=1R1+1R2+1R3        (I)

The resistor R4 and the combination R1,2,3 are in series. Hence their total resistance can be found out as,

  R1,2,3,4=R1,2,3+R4        (II)

The resistors R5, R6, are in parallel. So their net resistance is given by,

  R5,6=R5R6R5+R6        (III)

The resistor R7 and R5,6 are in series. Hence their total resistance is,

  R5,6,7=R5,6+R7        (IV)

The combinations R1,2,3,4 and R4,5,7 are in parallel. Thus, the equivalent resistance of the network can be found out as,

  Requivalent=R1,2,3,4R5,6,7R1,2,3,4+R5,6,7        (V)

Write the expression for the current in the EMF.

  I=εRequivalent        (VI)

Here, I is the current, and ε is the EMF.

Conclusion:

Substitute 2r for R1, 2r for R2, and r for R3 in equation (I) and solve for R1,2,3.

  1R1,2,3=12r+12r+1r1R1,2,3=42rR1,2,3=r2

Substitute r2 for R1,2,3, and 32r for R4 in equation (II) to find R1,2,3,4.

  R1,2,3,4=r2+32r=2r

Substitute r for R5 and R6 in equation (III) to find R5,6.

  R5,6=(r)(r)r+r=r2

Substitute r2 for R5,6, and 32r for R7 in equation (IV) to find R5,6,7.

  R5,6,7=r2+32r=2r

Substitute 2r for R1,2,3,4, and 2r for R5,6,7 in equation (V) to find Requivalent.

  Requivalent=(2r)(2r)2r+2r=r

Substitute r for R in equation (VI) to find I.

  I=εr

Therefore, the current in the EMF in the circuit shown in Figure 17.31 is ε/r_.

(b)

To determine

The current in the lower 32r resistor.

(b)

Expert Solution
Check Mark

Answer to Problem 71E

The current in the lower 32r resistor is ε/2r_.

Explanation of Solution

The lower 32r resistor is represented by R7 in the Figure 1.

Since the lower and upper branches of the network are connected in parallel, the potential difference across both the branches are same and equal to ε. Since R5,6 and R7 are in series, the current through both will be same. It can be computed as,

  I7=εR5,6,7        (VII)

Here, I7 is the current through R7.

Conclusion:

Substitute 2r for R5,6,7 in equation (VII) to find I7.

  I7=ε2r

Therefore, the current in the lower 32r resistor is ε/2r_.

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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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