PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 20, Problem 68P

(a)

To determine

The maximum current flows through the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 68P

The maximum current flows through the circuit is 45mA.

Explanation of Solution

Write an expression for the maximum current flows through the circuit.

  I0=εR                                                                                                                     (I)

Here, I0 is the maximum current flow through the circuit, ε is the emf and R is the resistance.

Conclusion:

Substitute 9.0V for ε and 200.0Ω for R in equation (I) to find I0.

  I0=9.0V200.0Ω=(45×103A)(1mA103A)=45mA

Thus, the maximum current flows through the circuit is 45mA.

(b)

To determine

The time taken for reaching the maximum current.

(b)

Expert Solution
Check Mark

Answer to Problem 68P

The time taken for reaching the maximum current is 1.0ms.

Explanation of Solution

Write an expression for the instantaneous current.

  I'=I0(1exp(tτ))                                                                                              (II)

Here, I' is the instantaneous current, t is the time and τ is the time constant.

Rearrange equation (II) to find t.

  t=τln(1I'I0)                                                                                                   (III)

Write an expression for the time constant.

  τ=LR                                                                                                                     (IV)

Here, L is the inductance and R is the resistance.

Rearrange the equation (III) using equation (IV).

  t=LRln(1I'I0)                                                                                                  (V)

Conclusion:

Substitute 0.30H for L, 200.0Ω for R and I0/2 for I'  in equation (V) to find t.

  t=0.30H200.0Ωln(1I0/2I0)=0.30H200.0Ωln(112)=(1.0×103s)(1ms103s)=1.0ms

Thus, the time taken for reaching the maximum current is 1.0ms.

(c)

To determine

The energy stored in the inductor and the rate at which energy is dissipated in the resistor.

(c)

Expert Solution
Check Mark

Answer to Problem 68P

The energy stored in the inductor is 76μJ and the rate at which energy is dissipated in the resistor is 0.10W.

Explanation of Solution

Write an expression for the energy stored in the inductor.

  U=12LI2                                                                                                          (VI)

Here, U is the energy.

The current is half of the maximum current. Thus, rewrite equation (VI).

  U=12L(I0/2)2                                                                                                   (VII)

Write an expression for the rate at which energy is converted in the resistor.

  P=I2R                                                                                                         (VIII)

The current is half of the maximum current. Thus, rewrite equation (VIII).

  P=(I0/2)2R                                                                                                    (IX)

Conclusion:

Substitute 0.30H for L, 200.0Ω for R and 45mA for I0  in equation (IX) to find U.

  U=12(0.30H)((45mA)(1A103mA)2)2=12(0.30H)(0.045m2)2=(76×106J)(1μJ106J)=76μJ

Substitute 0.30H for L, 200.0Ω for R and 45mA for I0  in equation (IX) to find P.

  P=((45mA)(1A103mA)/2)2(200.0Ω)=(45×103A/2)2(200.0Ω)=0.10W

Thus, the energy stored in the inductor is 76μJ and the rate at which energy is dissipated in the resistor is 0.10W.

(d)

To determine

The maximum current flows through the circuit and the time taken for reaching the maximum current.

(d)

Expert Solution
Check Mark

Answer to Problem 68P

The energy stored is 110kJ.

Explanation of Solution

Write an expression for equivalent resistance.

    Req=R+rL+r                                                                                                       (X)

Here, Req is the equivalent resistance, rL  is the internal resistance of inductor and r is the internal resistance of battery.

Write an expression for the maximum current flows through the circuit.

    Imax=εReq                                                       (XI)

Write an expression for the time taken for reaching the maximum current.

    t=LReqln(1I'I0)       (XII)

Conclusion:

Substitute 75Ω for rL, 20Ω for r and 200.0Ω for R in equation (X) to find Req.

    Req=75Ω+20Ω+200.0Ω=295Ω

Substitute 9.0V for ε and 295Ω for Req in equation (XI) to find Imax.

    I0=9.0V295Ω=(31×103A)(1mA103A)=31mA

Substitute 0.30H for L, 295Ω for Req and I0/2 for I'  in equation (XII) to find t.

    t=0.30H295Ωln(1I0/2I0)=0.30H295Ωln(112)=(0.70×103s)(1ms103s)=0.70ms

Thus, the maximum current flows through the circuit is 31mA and the time taken for reaching the maximum current is 0.70ms.

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