Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 118P

(a)

To determine

The comparison of total energy stored in the two capacitors.

(a)

Expert Solution
Check Mark

Answer to Problem 118P

The comparison of total energy stored in the two capacitors is UiUf=1+C2C1 .

Explanation of Solution

Formula Used:

The expression for the charge on the capacitor is given by,

  Q=CV0

The expression for the potential of the capacitor of the final stage.

  Q1C1=Q2C2

The expression for the total charge on the capacitor is given by,

  Q=Q1+Q2

The expression for the initial energies stored in the capacitor is given by,

  Ui=12C1V02 …… (I)

The expression for the final energies stored in the capacitor is given by,

  Uf=12Q12C1+12Q22C2 …… (II)

The expression for the charge Q1 is given by,

  Q1=C12C1+C2V0

The expression for charge Q2 is given by,

  Q2=C12C1+C2V0

Equation (II) can be re written as,

  Uf=12Q12C1+12Q22C2=12 ( C 1 2 C 1 + C 2 V 0 )2C1+12 ( C 1 C 2 C 1 + C 2 V 0 )2C2=12C12C1+C2V02

The expression for the ratios of initial and final energies is given by,

  UiUf=12C1V0212 C 1 2 C 1 + C 2 V02=1+C2C1

Conclusion:

Therefore, the comparison of total energy stored in the two capacitors is UiUf=1+C2C1 .

(b)

To determine

The current through R as a function of t .

(b)

Expert Solution
Check Mark

Answer to Problem 118P

The current through R as a function of t is I=V0Retτ .

Explanation of Solution

Formula Used:

The expression for the current passing through the circuit is given by,

  I=dq2dt

The expression Kirchhoff’s loop rule to circuit when the switch is at a is given by,

  q1C1IRq2C2=0q1C1Rdq2dtq2C2=0 ……. (III)

The expression for the law of the charge on the capacitors is given by,

  q1=Qq2=C1V0q2

Equation (III) can be re written as,

  q1C1Rdq2dtq2C2=0V0q2C1Rdq2dtq2C2=0V0=q2C1+Rdq2dt+q2C2 …… (IV)

Let the solution for the differential equation (IV) be,

  q2(t)=a+betτ

So,

  dq2(t)dt=ddt(a+b t τ )=bτetτ

Equation (IV) can be rewritten as,

  R(bτetτ)+(C1+C2C1C2)(a+betτ)=V0

Compare with solution of the equation.

  ( C 1 + C 2 C 1 C 2 )a=V0a=V0( C 1 C 2 C 1 + C 2 )=V0Ceq

And,

  R(bτe t τ )+( C 1 + C 2 C 1 C 2 )(be t τ )=0Rτ+C1+C2C1C2=0τ=R( C 1 C 2 C 1 + C 2 )=RCeq

From the initial condition,

  a+b=b=a=V0Ceq

The solution of the differential equation is,

  q2(t)=a+betτ=CeqV0+(C eqV0)etτ=CeqV0(1e t τ )

The current in the circuit is given by,

  I=ddt(C eqV0( 1 e t τ ))=C eqV0τetτ=V0Retτ

Conclusion:

Therefore, the current through R as a function of t is I=V0Retτ .

(c)

To determine

The energy delivered to resistor as a function of t .

(c)

Expert Solution
Check Mark

Answer to Problem 118P

The energy delivered to resistor as a function of t is P(t)=V02Retτ .

Explanation of Solution

Formula Used:

The current through R as a function of t is given by,

  I=V0Retτ

The expression for the energy dissipated through the resistor is given by,

  P(t)=I2R=( V 0 R e t τ )2R=V02Retτ

Conclusion:

Therefore, the energy delivered to resistor as a function of t is P(t)=V02Retτ .

(d)

To determine

The total energy dissipated.

(d)

Expert Solution
Check Mark

Answer to Problem 118P

The total energy dissipated is E=12(C1C2C1+C2)V02 .

Explanation of Solution

Formula Used:

The expression for the energy dissipated through the resistor is given by,

  P(t)=V02Retτ

The expression for the total energy dissipated through the resistor is given by,

  E=0P( t)dt=0 V 0 2 R e t τ =12V02Ceq=12( C 1 C 2 C 1 + C 2 )V02

Conclusion:

Therefore, the total energy dissipated is E=12(C1C2C1+C2)V02 .

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Chapter 25 Solutions

Physics for Scientists and Engineers

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