EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684651
Author: Katz
Publisher: VST
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Chapter 29, Problem 39PQ

(a)

To determine

Find the current through the emf device and each resistor in circuit 1.

(a)

Expert Solution
Check Mark

Answer to Problem 39PQ

The current through the Emf device and each resistor in circuit 1 is 0.917A_. The current for the given circuit resistance is 0.500A_ for R1, 0.250A_ for R2 and is 0.167A_ for R3.

Explanation of Solution

According to Kirchhoff’s junction rule, in any junction, the sum of the all the currents entering the junction equals the sum of all the currents exiting the junction.

Redraw the circuit 1 and labeled it as given below

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC, Chapter 29, Problem 39PQ , additional homework tip  1

In parallel circuit, voltage across all three resistors is same that means potential difference between node A and node B is same (VAVB=ε). Current will be divide equally in all three branches.

According to Ohm’s law,

  ε=IReq                                                                                                          (I)

Here, ε is the emf of the device, I is the total current and Req is the equivalent resistance of the circuit.

Rearrange the equation (I) in terms of total current

  I=εReq                                                                                                        (II)

Write the expression for equivalent resistance as.

    1Req=1R1+1R2+1R3

Rearrange the above expression.

    Req=(R1R2R3R1R2+R2R3+R1R3)                                                                         (III)

Write the expression for current I1.

    I1=εR1                                                                                                      (IV)

Here, I1 is current in first branch, and R1 is the resistance of first branch.

Write the expression for current I2

  I2=εR2                                                                                                    (V)

Here, I2 is current in second branch, and R2 is the resistance of second branch.

Write the expression for current I3

  I3=εR3                                                                                                     (VI)

Here, I3 is current in third branch, and R3 is the resistance of third branch.

Conclusion:

Substitute R Ω for R1, 2R Ω for R2 and 3R Ω for R3 in equation (III)

    Req=(R(2R)(3R)R(2R)+R(3R)+2R(3R))=(6R311R2)=(6R11)

Substitute 15 Ω for R. in above equation

    Req=(6(15Ω)11)=8.18 Ω

Substitute 8.18 Ω for Req and 7.50 V for ε in equation (II).

    I=7.50 V8.18 Ω=0.9168 A0.9167 A

Thus, the current in circuit 1 is 0.9167 A_.

Substitute 7.50 V for ε and 15 Ω for R1 in equation (IV).

    I1=7.50 V15.0 Ω=0.500 A

Substitute 7.50 V for ε and 30 Ω for R2 in equation (V).

    I2=7.50 V30.0 Ω=0.250 A

Substitute 7.50 V for ε and 45 Ω for R3 in equation (VI).

    I3=7.50 V45.0 Ω=0.167 A

Thus, the current through the Emf device and each resistor in circuit 1 is 0.9167 A_. The current for the given circuit resistance is 0.500 A_ for R1, 0.250 A_ for R2 and is 0.167 A_ for R3.

(b)

To determine

Find the current through the emf device and each resistor in circuit 2 refer to figure P29.28.

(b)

Expert Solution
Check Mark

Answer to Problem 39PQ

The current through the Emf device and each resistor in circuit 2 is 0.9167 A_. The current for the given circuit resistance is 0.500 A_ for R1, 0.250 A_ for R2 and is 0.167 A_ for R3.

Explanation of Solution

According to Kirchhoff’s junction rule, in any junction, the sum of the all the currents entering the junction equals the sum of all the currents exiting the junction.

Redraw the circuit 2 and labeled it as given below.

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC, Chapter 29, Problem 39PQ , additional homework tip  2

In parallel circuit 2, voltage across all three resistors is same that means potential difference between node a and node b, node c & node d and node e & node f are same (VAB=VCD=VEF=ε). Current will be divide equally in all three branches.

Write the expression for current I1 as.

    I1=VABR1                                                                                            (VII)

Here, I1 is the current in branch 1 , VAB is the voltage drop across ab and R1 is the resistance.

Write the expression for current I2 as.

  I2=VCDR2                                                                                            (VIII)

Here, I2 is the current in branch 2 , VCD is the voltage drop across cd and R2 is the resistance.

Write the expression for current I3 as.

  I3=VEFR3                                                                                              (IX)

Here, I3 is the current in branch 3 , VEF is the voltage drop across ef and R3 is the resistance.

Conclusion:

Substitute R Ω for R1, 2R Ω for R2 and 3R Ω for R3 in equation in equation (III)

    1Req=(2R(3R)+R(3R)+R(2R)R(2R)(3R)) =(6R2+3R2+2R26R3)=(11R26R3)=(116R)

Substitute 15 Ω for R. in above equation

    1Req=116(15Ω)=1190ΩReq=8.18 Ω

Substitute 8.18 Ω for Req and 7.50 V for ε in equation (II)

    I=7.50 V8.18 Ω=0.91680.9167 A

Thus, the current in circuit 1 is 0.9167 A_.

Substitute 7.50 V for VAB and 15 Ω for R1 in equation (VII).

    I1=7.50 V15.0 Ω=0.500 A

Substitute 7.50 V for VCD and 30 Ω for R2 in equation (VIII).

    I2=7.50 V30.0 Ω=0.250 A

Substitute 7.50 V for VEF and 45 Ω for R3 in equation (IX)

    I3=7.50 V45.0 Ω=0.167 A

Thus, the current for the given circuit resistance is 0.500 A_ for R1, 0.250 A_ for R2 and is 0.167 A_ for R3.

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Chapter 29 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

Ch. 29 - Prob. 5PQCh. 29 - Prob. 6PQCh. 29 - A real battery (modeled as an ideal emf device in...Ch. 29 - Prob. 8PQCh. 29 - Two circuits made up of identical ideal emf...Ch. 29 - Prob. 10PQCh. 29 - Prob. 11PQCh. 29 - Prob. 12PQCh. 29 - Eight real batteries, each with an emf of 5.00 V...Ch. 29 - Prob. 14PQCh. 29 - Prob. 15PQCh. 29 - Prob. 16PQCh. 29 - Prob. 17PQCh. 29 - Prob. 18PQCh. 29 - Prob. 19PQCh. 29 - An ideal emf device with emf is connected to two...Ch. 29 - Prob. 21PQCh. 29 - Prob. 22PQCh. 29 - Prob. 23PQCh. 29 - Prob. 24PQCh. 29 - Prob. 25PQCh. 29 - Prob. 26PQCh. 29 - Determine the currents through the resistors R2,...Ch. 29 - The emf devices in the circuits shown in Figure...Ch. 29 - Prob. 29PQCh. 29 - Prob. 30PQCh. 29 - Prob. 31PQCh. 29 - Prob. 32PQCh. 29 - Prob. 33PQCh. 29 - Prob. 34PQCh. 29 - A Figure P29.35 shows a combination of six...Ch. 29 - A Each resistor shown in Figure P29.36 has...Ch. 29 - Each resistor shown in Figure P29.36 has a...Ch. 29 - Prob. 38PQCh. 29 - Prob. 39PQCh. 29 - The emf in Figure P29.40 is 4.54 V. The...Ch. 29 - Figure P29.41 shows three resistors (R1 = 14.0 ,...Ch. 29 - Figure P29.42 shows five resistors and two...Ch. 29 - The emfs in Figure P29.43 are 1 = 6.00 V and 2 =...Ch. 29 - Prob. 44PQCh. 29 - Figure P29.45 shows five resistors connected...Ch. 29 - Figure P29.46 shows a circuit with a 12.0-V...Ch. 29 - Two ideal emf devices are connected to a set of...Ch. 29 - Two ideal emf devices are connected to a set of...Ch. 29 - Three resistors with resistances R1 = R/2 and R2 =...Ch. 29 - Prob. 51PQCh. 29 - Prob. 52PQCh. 29 - Prob. 53PQCh. 29 - Prob. 55PQCh. 29 - At time t = 0, an RC circuit consists of a 12.0-V...Ch. 29 - A 210.0- resistor and an initially uncharged...Ch. 29 - Prob. 58PQCh. 29 - A real battery with internal resistance 0.500 and...Ch. 29 - Figure P29.60 shows a simple RC circuit with a...Ch. 29 - Prob. 61PQCh. 29 - Prob. 62PQCh. 29 - Prob. 63PQCh. 29 - Ralph has three resistors, R1, R2, and R3,...Ch. 29 - Prob. 65PQCh. 29 - An ideal emf device is connected to a set of...Ch. 29 - Prob. 67PQCh. 29 - An ideal emf device (24.0 V) is connected to a set...Ch. 29 - Prob. 69PQCh. 29 - What is the equivalent resistance between points a...Ch. 29 - A capacitor with initial charge Q0 is connected...Ch. 29 - Prob. 73PQCh. 29 - Prob. 74PQCh. 29 - Prob. 75PQCh. 29 - Prob. 76PQCh. 29 - Figure P29.77 shows a circuit with two batteries...Ch. 29 - In the RC circuit shown in Figure P29.78, an ideal...Ch. 29 - Prob. 79PQCh. 29 - Calculate the equivalent resistance between points...Ch. 29 - In Figure P29.81, N real batteries, each with an...Ch. 29 - Prob. 82PQCh. 29 - Prob. 83PQCh. 29 - Prob. 84PQCh. 29 - Figure P29.84 shows a circuit that consists of two...Ch. 29 - Prob. 86PQCh. 29 - Prob. 87PQCh. 29 - Prob. 88PQCh. 29 - Prob. 89PQCh. 29 - Prob. 90PQCh. 29 - Prob. 91PQCh. 29 - Prob. 92PQCh. 29 - Prob. 93PQCh. 29 - Prob. 94PQCh. 29 - Prob. 95PQ
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