Physics for Scientists and Engineers, Volume 1, Chapters 1-22
Physics for Scientists and Engineers, Volume 1, Chapters 1-22
8th Edition
ISBN: 9781439048382
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 27.63AP

A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit shown in Figure P26.37, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is then closed, and the circuit comes to equilibrium. In terms of Q and C, find (a) the final potential difference between the plates of each capacitor, (b) the charge on each capacitor, and (c) the final energy stored in each capacitor. (d) Find the internal energy appearing in the resistor.

Figure P26.37

Chapter 27, Problem 27.63AP, A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit

(a)

Expert Solution
Check Mark
To determine
The final potential difference between the plates of each capacitor.

Answer to Problem 27.63AP

The final potential difference between the plates of each capacitor is Q4C .

Explanation of Solution

Given information: The value of capacitor into the circuit is C and initial uncharged capacitor of capacitance is 3C , the charge placed on a capacitor is Q .

Explanation:

Formula to calculate the equivalent capacitance of the system when they are connected in parallel.

Ceq=C1+C2 (1)

Here,

Ceq is the equivalent capacitance of the system when they are connected in parallel.

C1 is the value of capacitor into the circuit.

C2 is the initial uncharged capacitor of capacitance.

Substitute C for C1 , 3C for C2 in equation (1) to find Ceq ,

CP=C+3C=4C

Thus, the equivalent capacitance of the system when they are connected in parallel is 4C .

Formula to calculate the final potential difference between the plates of each capacitor.

Vf=QCeq (2)

Here,

Vf is the final potential difference between the plates of each capacitor.

Q is the charge placed on a capacitor.

Substitute 4C for Ceq in equation (2) to find Vf ,

Vf=Q4C

Thus, the final potential difference between the plates of each capacitor is Q4C .

Conclusion:

Therefore, the final potential difference between the plates of each capacitor is Q4C .

(b)

Expert Solution
Check Mark
To determine
The charge on each capacitor.

Answer to Problem 27.63AP

The charge on capacitor C is Q4 and charge on capacitor 3C is 3Q4 .

Explanation of Solution

Given information: The value of capacitor into the circuit is C and initial uncharged capacitor of capacitance is 3C , the charge placed on a capacitor is Q .

Explanation:

Formula to calculate the charge placed on a capacitor C .

QC=C1Vf (3)

Here,

QC is the charge placed on a capacitor C .

Substitute C for C1 in equation (3) to find QC ,

QC=C×Q4C=Q4

Thus, the charge placed on a capacitor C is Q4 .

Formula to calculate the charge placed on a capacitor 3C .

Q3C=C2Vf (4)

Here,

Q3C is the charge placed on a capacitor 3C .

Substitute 3C for C2 in equation (4) to find Q3C ,

Q3C=3C×Q4C=3Q4

Thus, the charge placed on a capacitor 3C is 3Q4 .

Conclusion:

Therefore, the charge on capacitor C is Q4 and charge on capacitor 3C is 3Q4 .

(c)

Expert Solution
Check Mark
To determine
The final energy stored in each capacitor.

Answer to Problem 27.63AP

The energy stored in the capacitor C is Q232C , energy stored in the capacitor 3C is 3Q232C .

Explanation of Solution

Given information: The value of capacitor into the circuit is C and initial uncharged capacitor of capacitance is 3C , the charge placed on a capacitor is Q .

Explanation:

Formula to calculate the energy stored in the capacitor C .

EC=12C1Vf2 (5)

Here,

EC is the energy stored in the capacitor C .

Substitute Q4C for Vf , C for C1 in equation (5) to find EC ,

EC=12(C)×(Q4C)2=Q232C

Thus, the energy stored in the capacitor C is Q232C .

Formula to calculate the energy stored in the capacitor 3C .

E3C=12C2Vf2 (6)

Here,

E3C is the energy stored in the capacitor 3C .

Substitute Q4C for Vf , 3C for C2 in equation (6) to find E3C ,

E3C=12(3C)×(Q4C)2=3Q232C

Thus, the energy stored in the capacitor 3C is 3Q232C .

Conclusion:

Therefore, the energy stored in the capacitor C is Q232C , energy stored in the capacitor 3C is 3Q232C .

(d)

Expert Solution
Check Mark
To determine
The internal energy appearing in the resistor.

Answer to Problem 27.63AP

The internal energy appearing in the resistor is 3Q28C .

Explanation of Solution

Given information: The value of capacitor into the circuit is C and initial uncharged capacitor of capacitance is 3C , the charge placed on a capacitor is Q .

Explanation:

Write the expression for the original energy in the resistor.

Eoriginal=Q22C (7)

Here,

Eoriginal is the original energy in the resistor.

Formula to calculate the internal energy appearing in the resistor.

Er=Eoriginal(EC+E3C)

Here,

Er is the internal energy appearing in the resistor.

Substitute Q22C for Eoriginal , Q232C for EC , 3Q232C for E3C in equation (7) to find Er ,

Er=Q22C(Q232C+3Q232C)=Q28C

Thus, the internal energy appearing in the resistor is 3Q28C .

Conclusion:

Therefore, the internal energy appearing in the resistor is 3Q28C .

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Chapter 27 Solutions

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

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