The potential difference across each capacitor.

Question

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Answer 1

Question

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

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Answer 2

Question

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

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Answer 3

Question

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

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Answer 4

Question

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

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Answer 5

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

Answer 6

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

Answer 7

Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

Answer 8

(a)

Expert Solution

Answer to Problem 54P

V=4.50 V

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Potential difference

V=QC

Where,

Q is the charge

C is the capacitance

Calculation:

Charge on the capacitor when they are initially connected in series can be calculated as:

Q=CeqVQ=( C 1 C 2 C 1 + C 2 )VQ=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))×12Q=( 4 8× 10 −12 ( 16 × 10 −6 ))×12Q=3×10−6 ×12Q=36×10−6 CQ=36 μC

When they are connected in parallel, net charge is:

Qnet=2×36 Qnet=72 μC

Equivalent capacitor when connected in parallel is

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

In parallel combination, potential difference is same across all the capacitors.

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Conclusion:

Potential difference across each capacitor is 4.5 V

Answer 13

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

Answer 14

(b)

To determine

To Find: The energy stored in capacitor before disconnection and after reconnection.

Answer 15

(b)

Expert Solution

Answer to Problem 54P

Ei=216×10−6 J

Ef=162×10−6 J

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Capacitor 1, C1=4 μF = 4 ×10−6 F

Capacitor 2, C2=12 μF = 12 ×10−6 F

Potential difference, V=12 V

Formula used:

Energy stored

E=12CV2

Where,

C is the capacitance

V is the potential difference

Calculation:

Equivalent capacitance when capacitors are connected in series:

Ceq=( C 1 C 2 C 1 + C 2 )=( 4 × 10 −6 ×12 × 10 −6 ( 4 × 10 −6 )+( 12 × 10 −6 ))=4 8× 10 −12( 16 × 10 −6 )=3×10−6 C

Initially energy stored in capacitor is:

E=12CeqV2E=12×3×10−6 ×(12)2Ei=216×10−6 J

Equivalent capacitance when capacitors connected in parallel:

Ceq'=C1+C2Ceq'=(4 × 10 −6 )+(12 × 10 −6)Ceq'=16 ×10−6 CCeq'=16 μC

Potential difference can be calculated as:

V'=Q netC eq'V'=72 × 10 −616× 10 −6V'=4.5 V

Final energy stored in capacitor is:

Ef=12Ceq'V'2Ef=12×16×10−6 ×(4.5)2Ef=162×10−6 J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×10−6 J and after reconnection is Ef=162×10−6 J .

Answer 20

Ch. 24 - Prob. 1P Ch. 24 - Prob. 2P Ch. 24 - Prob. 3P Ch. 24 - Prob. 4P Ch. 24 - Prob. 5P Ch. 24 - Prob. 6P Ch. 24 - Prob. 7P Ch. 24 - Prob. 8P Ch. 24 - Prob. 9P Ch. 24 - Prob. 10P

The potential difference across each capacitor.

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To Find: The potential difference across each capacitor.

Answer to Problem 54P

Explanation of Solution

To Find: The energy stored in capacitor before disconnection and after reconnection.

Answer to Problem 54P

Explanation of Solution

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