EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
1st Edition
ISBN: 9780100546714
Author: Katz
Publisher: YUZU
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Chapter 33, Problem 23PQ

In the LC circuit in Figure 33.11, the inductance is L = 19.8 mH and the capacitance is C = 19.6 mF. At some moment, UB = UE= 17.5 mJ. a. What is the maximum charge stored by the capacitor? b. What is the maximum current in the circuit? c. At t = 0, the capacitor is fully charged. Write an expression for the charge stored by the capacitor as a function of lime. d. Write an expression for the current as a function of time.

(a)

Expert Solution
Check Mark
To determine

The maximum charge stored by the capacitor.

Answer to Problem 23PQ

The maximum charge stored by the capacitor is 3.7×102C.

Explanation of Solution

Write the expression to calculate the total; energy.

    ET=UB+UE                                         (I)

Here, ET is total energy, UB is energies stored in magnetic field, and UE is energy stored in electric field.

Write the expression to calculate the maximum charge.

    ET=Q2max2C                                                (II)

Here, Qmax is maximum charge, and C is capacitance.

Conclusion:

Substitute 17.5mJ for UB and UE in equation (I) to calculate ET.

    ET=17.5mJ×(1J103mJ)+17.5mJ×(1J103mJ)=17.5×103J+17.5×103J=35×103J

Substitute 19.6mF for C, and 35×103J for ET in equation (II) to calculate Qmax.

    35×103J=Q2max2×19.6mF×(1F103mF)Q2max=1372×106C2Qmax=1372×106C2Qmax=3.7×102C

Therefore, the maximum charge stored by the capacitor is 3.7×102C.

(b)

Expert Solution
Check Mark
To determine

The maximum current in the circuit.

Answer to Problem 23PQ

The maximum current in the capacitor is 1.88A.

Explanation of Solution

Write the expression to calculate the maximum current.

    ET=12LI2max                                                    (III)

Here, Imax is maximum current, and L is inductance.

Conclusion:

Substitute 19.8×103H for L, and 35×103J for ET in Equation (III) to calculate Imax.

    35×103J=12×19.8×103H×I2maxI2max=3.53A2Imax=3.53A2Imax=1.88A

Therefore, the maximum current in the capacitor is 1.88A.

(c)

Expert Solution
Check Mark
To determine

The expression for the charge stored by the capacitor as function of time.

Answer to Problem 23PQ

The expression for the charge stored by the capacitor as function of time is Q(t)=3.7×102cos(50.8t).

Explanation of Solution

Write the expression to calculate the angular frequency.

    ω=1LC                                                             (IV)

Here, ω is angular frequency.

Write the expression for the charge stored by the capacitor as function of time Q(t).

    Q(t)=Qmaxcos(ωt+ϕ)                                       (V)

Here, Qmax is maximum charge.

Conclusion:

Substitute 19.8×103H for L, and 19.6×103F for C in Equation (IV) to calculate ω.

    ω=119.8×103H×19.6×103F=1388.08×106radian/s=50.8radian/s

Substitute 0 for ϕ, 50.8radian/s for ω, and 3.7×102C for Qmax in Equation (V) to calculate Q(t).

    Q(t)=3.7×102Ccos(50.8radian/s×t+0)=3.7×102cos(50.8t)

Therefore, the expression for the charge stored by the capacitor as function of time is Q(t)=3.7×102cos(50.8t).

(d)

Expert Solution
Check Mark
To determine

The expression for the current as function of time.

Answer to Problem 23PQ

The expression for the current as function of time is I(t)=1.88cos(50.8t).

Explanation of Solution

Write the expression for current as function of time I(t).

    I(t)=Imaxcos(ωt+ϕ)                                       (VI)

Here, Imax is maximum current.

Conclusion:

Substitute 0 for ϕ, 50.8radian/s for ω, and 1.88A for Imax in equation (VI) to calculate I(t)

    I(t)=1.88Acos(50.8radian/s×t+0)=1.88cos(50.8t)

Therefore, the expression for the current as function of time is I(t)=1.88cos(50.8t).

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Chapter 33 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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