Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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Chapter 29, Problem 43PQ

The emfs in Figure P29.43 are ε1 = 6.00 V and ε2 = 12.0 V. The resistances are R1 = 15.0 Ω, R2 = 30.0 Ω, R3 = 45.0 Ω, and R4 = 60.0 Ω. Find the current in each resistor when the switch is

  1. a. open and
  2. b. closed.

Chapter 29, Problem 43PQ, The emfs in Figure P29.43 are 1 = 6.00 V and 2 = 12.0 V. The resistances are R1 = 15.0 , R2 = 30.0 ,

(a)

Expert Solution
Check Mark
To determine

The current in each resistor when the switch is open.

Answer to Problem 43PQ

The current flowing through the resistor R1 is 0.133A, R2 is 0.133A, R3 is 0.133A and R4 is 0A.

Explanation of Solution

As the switch is kept open; the current in the wire EFAB will be zero as the circuit is disconnected.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 29, Problem 43PQ , additional homework tip  1

The resistors in the loop BCDEB are connected in series. The current flowing in all the elements in a series circuit is constant.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 29, Problem 43PQ , additional homework tip  2

Write the expression for the equivalent resistance in the loop BCDEB as.

  Req=R1+R2+R3                                                                                                     (I)

Here, Req is the equivalent series resistance, R1 is the resistance of first element, R2 is the resistance of second element and R3 is the resistance of third element.

Write the expression for the current in the loop BCDEB as.

    I=ε2Req                                                                                                                    (II)

Here, I is the current in the loop BCDEB and ε2 is the Emf of the battery.

No current flows through the resistor R4 as the switch is kept open and therefore the value of the current I0 in the loop ABEFA is zero.

Conclusion:

Substitute 15Ω for R1 , 30Ω for R2 and 45Ω for R3 in equation (I).

    Req=15Ω+30Ω+45Ω=90Ω

Substitute 12V for ε2 and 90Ω for Req in equation (II).

  I=12V90Ω=0.133A

Thus, the current flowing through the resistor R1 is 0.133A, R2 is 0.133A, R3 is 0.133A and R4 is 0A.

(b)

Expert Solution
Check Mark
To determine

The current in each resistor when the switch is closed.

Answer to Problem 43PQ

The current flowing through the resistor R1 is 0.179A, R2 is 0.124A, R3 is 0.124A and R4 is 0.0552A.

Explanation of Solution

The switch is closed, the current starts flowing through all the resistors in the circuit.

The circuit diagram for the closed switch is shown below.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 29, Problem 43PQ , additional homework tip  3

Write the expression for the Kirchhoff’s voltage law in loop ABCDEFA.

    ε1I'R3+ε2I'R2+IoR4=0                                                                               (III)

Here, I' is the current flowing in loop BCDEB and Io is the current flowing in loop ABEFA, R4 is the resistance of fourth element and ε1 is the Emf of battery.

Write the expression for the Kirchhoff’s voltage law in loop ABEFA.

  ε1+(Io+I')R1+IoR4=0                                                                                     (IV)

Write the expression for the current through resistor R1.

  IR1=Io+I'                                                                                                               (V)

Here, IR1 is the current in resistor R1.

Write the expression for the current through resistor R2.

  IR2=I'                                                                                                                    (VI)

Here, IR2 is the current in resistor R2.

Write the expression for the current through resistor R3.

  IR3=I'                                                                                                                   (VII)

Here, IR3 is the current in resistor R3.

Write the expression for the current through resistor R4.

  IR4=Io                                                                                                                 (VIII)

Here, IR4 is the current in resistor R4.

Conclusion:

Substitute 6V for ε1 , 45Ω for R312V for ε2 , 30Ω for R2 and 60Ω for R4 in equation (III).

  (6V)I'(45Ω)+(12V)I'(30Ω)+Io(60Ω)=075I'60Io=6

Rearrange the above equation for I' .

  I'=6+60Io75                                                                                                           (IX)

Substitute 6V for ε1, 15Ω for R1 and 60Ω for R4 in equation (IV).

  (6V)+(Io+I')(15Ω)+Io(60Ω)=075Io+15I'=6

Substitute 6+60Io75 for I' in above equation.

  75Io+15(6+60Io75)=675Io+12I'=245

Rearrange the above expression for Io.

    Io=0.0552A

Substitute 0.0552A for Io in equation (IX).

  I'=6+((60)(0.0552A))75=0.124A

Substitute 0.0552A for Io and 0.124A for I' in equation (V).

  IR1=(0.0552A)+(0.124A)=0.179A

Substitute 0.124A for I' in equation (VI).

  IR2=0.124A

Substitute 0.124A for I' in equation (VII).

  IR3=0.124A

Substitute 0.0552A for Io in equation (V).

  IR4=0.0552A

Thus, the current flowing through the resistor R1 is 0.179A, R2 is 0.124A, R3 is 0.124A and R4 is 0.0552A.

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Chapter 29 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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