EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 24, Problem 54P

(a)

To determine

To Find: The potential difference across each capacitor.

(a)

Expert Solution
Check Mark

Answer to Problem 54P

  V=4.50V

Explanation of Solution

Given:

Capacitor 1, C1=4μF = 4×106

Capacitor 2, C2=12μF = 12×106

Potential difference, V=12V

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=( 48× 10 12 ( 16× 10 6 ))×12Q=3×106×12Q=36×106CQ=36μC

When they are connected in parallel, net charge is:

  Qnet=2×36Qnet=72μC

Equivalent capacitor when connected in parallel is

  Ceq'=C1+C2Ceq'=(4× 10 6)+(12× 10 6)Ceq'=16×106CCeq'=16μC

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'=4.5V

Conclusion:

Potential difference across each capacitor is 4.5V

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 54P

  Ei=216×106J

  Ef=162×106J

Explanation of Solution

Given:

Capacitor 1, C1=4μF = 4×106

Capacitor 2, C2=12μF = 12×106

Potential difference, V=12V

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 ))=48× 10 12( 16× 10 6 )=3×106C

Initially energy stored in capacitor is:

  E=12CeqV2E=12×3×106×(12)2Ei=216×106J

Equivalent capacitance when capacitors connected in parallel:

  Ceq'=C1+C2Ceq'=(4× 10 6)+(12× 10 6)Ceq'=16×106CCeq'=16μC

Potential difference can be calculated as:

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

Final energy stored in capacitor is:

  Ef=12Ceq'V'2Ef=12×16×106×(4.5)2Ef=162×106J

Conclusion:

Energy stored in capacitor before disconnection is Ei=216×106J and after reconnection is Ef=162×106J .

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Chapter 24 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 24 - Prob. 11PCh. 24 - Prob. 12PCh. 24 - Prob. 13PCh. 24 - Prob. 14PCh. 24 - Prob. 15PCh. 24 - Prob. 16PCh. 24 - Prob. 17PCh. 24 - Prob. 18PCh. 24 - Prob. 19PCh. 24 - Prob. 20PCh. 24 - Prob. 21PCh. 24 - Prob. 22PCh. 24 - Prob. 23PCh. 24 - Prob. 24PCh. 24 - Prob. 25PCh. 24 - Prob. 26PCh. 24 - Prob. 27PCh. 24 - Prob. 28PCh. 24 - Prob. 29PCh. 24 - Prob. 30PCh. 24 - Prob. 31PCh. 24 - Prob. 32PCh. 24 - Prob. 33PCh. 24 - Prob. 34PCh. 24 - Prob. 35PCh. 24 - Prob. 36PCh. 24 - Prob. 37PCh. 24 - Prob. 38PCh. 24 - Prob. 39PCh. 24 - Prob. 40PCh. 24 - Prob. 41PCh. 24 - Prob. 42PCh. 24 - Prob. 43PCh. 24 - Prob. 44PCh. 24 - Prob. 45PCh. 24 - Prob. 46PCh. 24 - Prob. 47PCh. 24 - Prob. 48PCh. 24 - Prob. 49PCh. 24 - Prob. 50PCh. 24 - Prob. 51PCh. 24 - Prob. 52PCh. 24 - Prob. 53PCh. 24 - Prob. 54PCh. 24 - Prob. 55PCh. 24 - Prob. 56PCh. 24 - Prob. 57PCh. 24 - Prob. 58PCh. 24 - Prob. 59PCh. 24 - Prob. 60PCh. 24 - Prob. 61PCh. 24 - Prob. 62PCh. 24 - Prob. 63PCh. 24 - Prob. 64PCh. 24 - Prob. 65PCh. 24 - Prob. 66PCh. 24 - Prob. 67PCh. 24 - Prob. 68PCh. 24 - Prob. 69PCh. 24 - Prob. 70PCh. 24 - Prob. 71PCh. 24 - Prob. 72PCh. 24 - Prob. 73PCh. 24 - Prob. 74PCh. 24 - Prob. 75PCh. 24 - Prob. 76PCh. 24 - Prob. 77PCh. 24 - Prob. 78PCh. 24 - Prob. 79PCh. 24 - Prob. 80PCh. 24 - Prob. 81PCh. 24 - Prob. 82PCh. 24 - Prob. 83PCh. 24 - Prob. 84PCh. 24 - Prob. 85PCh. 24 - Prob. 86PCh. 24 - Prob. 87PCh. 24 - Prob. 88PCh. 24 - Prob. 89PCh. 24 - Prob. 90PCh. 24 - Prob. 91PCh. 24 - Prob. 92PCh. 24 - Prob. 93PCh. 24 - Prob. 94P
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