Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 21, Problem 92P

(a)

To determine

The impedance of the inductor in the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 92P

The impedance of the inductor in the circuit is 33.8Ω.

Explanation of Solution

Write the expression for impedance of the RL series circuit.

  Z=R2+XL2                                                                                  (I)

Here, Z is the impedance, R is the resistance, and XL is the inductive reactance.

Write the expression for inductive reactance.

  XL=ωL                                                                                           (II)

Here, Physics, Chapter 21, Problem 92P  is the angular frequency and L is the inductance.

Conclusion:

Substitute equation (II) in the equation (I).

  Z=R2+ω2L2                                                                                (III)

Write the equation for the induced emf of the ac source.

  ε(t)=(286V)sin[(390rad/s)t]

Here, the angular frequency of the ac mains is 390rad/s.

Substitute 390rad/s for ω, 30.0Ω for R, and 40.0mH for L in equation (III) to find Z.

  Z=(30.0Ω)2+(390rad/s)2[(40.0mH)(103H1mH)]2=33.8Ω

Therefore, the impedance of the inductor in the circuit is 33.8Ω.

(b)

To determine

The peak and rms voltage across the inductor.

(b)

Expert Solution
Check Mark

Answer to Problem 92P

The peak voltage across the inductor is 286V and rms voltage across the inductor is 202V.

Explanation of Solution

The peak voltage across the inductor including its internal resistance is equal to the peak source voltage,

  VL=εm                                                                                         (IV)

Here, VL is the voltage across the inductor and εm is the peak source voltage.

Write the expression for rms voltage across the inductor.

  Vrms=VL2                                                                                       (V)

Here, Vrms is the rms voltage.

Conclusion:

Substitute 286V for εm in equation (IV) to find VL.

  VL=286V

Substitute 286V for VL in the equation (V) to find Vrms

  Vrms=286V2=202.3V=202V

Therefore, the peak voltage across the inductor is 286V and rms voltage across the inductor is 202V.

(c)

To determine

The peak current in the circuit.

(c)

Expert Solution
Check Mark

Answer to Problem 92P

The peak current in the circuit is 8.46A.

Explanation of Solution

Write the expression for peak current.

  I=εmZ                                                                                                 (VI)

Here, I is the peak current.

Conclusion:

Substitute 286V for εm and 33.8Ω for Z in equation (VI) to find I.

  I=286V33.8Ω=8.46A

Therefore, the peak current in the circuit is 8.46A.

(d)

To determine

The average power dissipated in the circuit.

(d)

Expert Solution
Check Mark

Answer to Problem 92P

The average power dissipated in the circuit is 1.07kW.

Explanation of Solution

Write the expression average power dissipated in the circuit.

  Pavg=Irmsεrmscosϕ                                                                             (VII)

Here, Pavg is the average power dissipated in the circuit, Irms is the rms current, ϕ is the phase angle, and εrms is the rms voltage.

Write the expression for the phase angle.

  cosϕ=RZ                                                                                               (VIII)

Write the expression for the rms current.

  Irms=I2                                                                                                (IX)

Conclusion:

Substitute the equation (VIII) and (IX) in the equation (VII).

  Pavg=I2εrmsRZ

Substitute 8.46A for I, 202V for εrms, 30.0Ω for R, and 33.8Ω for Z to find Pavg.

  Pavg=8.46A2(202V)30.0Ω33.8Ω=1072.8W=1.07×103W(103kW1W)=1.07kW

The average power dissipated in the circuit is 1.07kW.

(e)

To determine

Derive the expression for the current through the inductor.

(e)

Expert Solution
Check Mark

Answer to Problem 92P

The expression for the current through the inductor is

i(t)=(8.46A)sin[(390rad/s)t0.480rad].

Explanation of Solution

Write the expression for current through the inductor lags the voltage across it.

  i(t)=Ipeaksin(ωtϕ) (X)

Rewrite the expression for the phase angle from part (d) in the equation (VIII).

  ϕ=cos1(RZ) (XI)

Conclusion:

Substitute 30.0Ω for R, and 33.8Ω for Z to find ϕ in equation (XI).

  ϕ=cos1(30.0Ω33.8Ω)=0.480rad

Substitute 8.46A for I, 390rad/s for ω, and 0.480rad for ϕ in equation (X) to find i(t).

  i(t)=(8.46A)sin[(390rad/s)t0.480rad]

Therefore, the expression for the current through the inductor is

i(t)=(8.46A)sin[(390rad/s)t0.480rad].

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Chapter 21 Solutions

Physics

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