Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 27, Problem 28PQ

(a)

To determine

The equivalent capacitance for the network given.

(a)

Expert Solution
Check Mark

Answer to Problem 28PQ

The equivalent capacitance of the combination is 1.71μF.

Explanation of Solution

Write the formula for equivalent capacitance of capacitors in series.

    1CS=1C1+1C2+1C3

Here, CS is the equivalent capacitance of the series combination of capacitors.

Write the formula for equivalent capacitance of capacitors in parallel.

    CP=C1+C2+C3+

Here, CP is the equivalent capacitance of the parallel combination of capacitors.

The capacitor CA and CB are in parallel to each other.

Write the formula for the resultant of CA and CB.

  CAB=CA+CB

Here, CAB is the resultant capacitance of CA and CB.

The capacitance CAB and CC are in series.

Write the formula for the equivalent capacitance of the given combination.

    1Ce=1CAB+1CC=1CA+CB+1CC

Here, Ce is the equivalent capacitance of the combination.

Solve the above equation to get expression for Ce.

    Ce=[1CA+CB+1CC]1

Conclusion:

Substitute 1.00μF for CA, 2.00μF for CB, 4.00μF for CC to determine Ce.

    Ce=[11.00μF+2.00μF+14.00μF]1=1.71μF

The equivalent capacitance of the combination is 1.71μF.

(b)

To determine

The charge stored in each of the capacitors.

(b)

Expert Solution
Check Mark

Answer to Problem 28PQ

The charge on capacitor CA is 5.14μC. The charge on capacitor CB is 10.3μC. The charge on capacitor CC is 15.4μC.

Explanation of Solution

The capacitors in series CAB and CC will be same and equal to the charge on equivalent capacitor of the combination.

Write the formula for the charge on equivalent capacitor.

    Qe=CeVs

Here, Ce is the equivalent capacitance of the combination, Qe is the charge on the equivalent capacitor, and Vs is the voltage provided by the supply.

Write the formula for the voltage across capacitor CA and CB.

    VAB=QABCAB=QABCA+CB

Here, VAB is the voltage across capacitor CA and CB, QAB is the charge on equivalent capacitance of capacitor CA and CB, and CAB is the equivalent capacitance of capacitor CA and CB.

Write the formula for the charge across CA.

    QA=CAVAB

Here, QA is the charge across CA.

Write the formula for the charge across CB.

    QB=CBVAB

Here, QB is the charge across CB.

Conclusion:

Substitute 1.71μF for Ce, 9.00V for Vs to determine Qe.

    Qe=(1.71μF)(9.00V)=15.4μC

Substitute 15.4μC for QAB, 1.00μF for CA, 2.00μF for CB to determine VAB.

    VAB=15.4μC1.00μF+2.00μF=5.14V

Substitute 5.14V for VAB, 1.00μF for CA to determine QA.

    QA=(1.00μF)(5.14V)=5.14μC

Substitute 5.14V for VAB, 2.00μF for CB to determine QB.

    QB=(2.00μF)(5.14V)=10.3μC

The charge on capacitor CA is 5.14μC. The charge on capacitor CB is 10.3μC. The charge on capacitor CC is 15.4μC.

(c)

To determine

The voltage across each of the capacitors.

(c)

Expert Solution
Check Mark

Answer to Problem 28PQ

The voltage across CA and CB is 5.14V. The voltage across CC is 3.86V.

Explanation of Solution

The capacitors in series CAB and CC will be same and equal to the charge on equivalent capacitor of the combination.

Write the formula for the charge on equivalent capacitor.

    Qe=CeVs

Here, Ce is the equivalent capacitance of the combination, Qe is the charge on the equivalent capacitor, and Vs is the voltage provided by the supply.

Write the formula for the voltage across capacitor CA and CB.

    VAB=QABCAB=QABCA+CB

Here, VAB is the voltage across capacitor CA and CB, QAB is the charge on equivalent capacitance of capacitor CA and CB, and CAB is the equivalent capacitance of capacitor CA and CB.

Write the formula for the voltage across CC.

    VC=QCCC

Here, QC is the charge across CC, VC is the voltage across CC.

Conclusion:

Substitute 1.71μF for Ce, 9.00V for Vs to determine Qe.

    Qe=(1.71μF)(9.00V)=15.4μC

Substitute 15.4μC for QAB, 1.00μF for CA, 2.00μF for CB to determine VAB.

    VAB=15.4μC1.00μF+2.00μF=5.14V

Substitute 15.4μC for QC, 4.00μF for CC to determine VC.

    VC=15.4μC4.00μF=3.86V

The voltage across CA and CB is 5.14V. The voltage across CC is 3.86V.

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Chapter 27 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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