Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 21, Problem 69P

(a)

To determine

The charge on the capacitor C1 in the electric circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 69P

The charge on the capacitor C1 in the electric circuit is 222μC_.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 69P , additional homework tip  1

The switch is closed in the electric circuit and the current exists in a simple series circuit as shown in figure 1.

Write the expression for the power delivered to the resistor.

    P=I2R2        (I)

Here, P is the power delivered to the resistor, I is the current, R2 is the resistance.

Use equation (I) to solve for I.

    I=PR2        (II)

Write the expression for the potential difference across the resistor R1.

    ΔV1=IR1        (III)

Here, ΔV1 is the potential difference across the resistor, R1 is the resistance.

Write the expression for the charge on the capacitor C1.

    Q1=C1ΔV1        (IV)

Here, C1 is the capacitance, Q1 is the charge on the C1 .

Conclusion:

Substitute 2.40W for P, 7.00kΩ for R2 in equation (II) to find I.

    I=2.40W7.00kΩ×103Ω1kΩ=18.5×103A=18.5mA

Substitute 18.5mA for I, 4.00kΩ for R1 in equation (III) to find ΔV1.

    ΔV1=(18.5mA×103A1mA)(4.00kΩ×103Ω1kΩ)=74.1V

Substitute 3.00μF for C1, 74.1V for ΔV1 in equation (IV) to find Q1.

    Q1=(3.00μF×106F1μF)(74.1V)=222×106C=222μC

Therefore, the charge on the capacitor C1 in the electric circuit is 222μC_.

(b)

To determine

The amount of charge on the capacitor C2 is changed.

(b)

Expert Solution
Check Mark

Answer to Problem 69P

The amount of charge on the capacitor C2 is changed is 444μC_.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 69P , additional homework tip  2

Consider the switch is closed to find the emf of the battery and the charge in the capacitor C2.

Write the expression for the potential difference across R2.

    ΔV2=IR2        (V)

Write the expression for the charge on the capacitor C2.

    Q2=C2ΔV2        (VI)

Write the expression for the emf of the battery.

    V=IReq        (VII)

Here, V is the emf of the battery, Req is the equivalent resistance.

Write the expression for Req.

    Req=R1+R2        (VIII)

Use equation (VIII) in (VII) to solve for V.

    V=I(R1+R2)        (IX)

Here, V is the emf of the battery.

After the switch is opened, no current exists. The potential difference across each resistor is zero. The emf of the battery appears across both capacitors.

Write the expression for the new charge on the C2.

    Q=C2V        (X)

Write the expression for the amount of the charge on the capacitor is changed C2.

    ΔQ=QQ2        (XI)

Here, ΔQ is the change in the charge on the C2.

Conclusion:

Substitute 18.5mA for I, 7.0kΩ for R2 in equation (V) to find ΔV2.

    ΔV2=(18.5mA×103A1mA)(7.00kΩ×103Ω1kΩ)=130V

Substitute 6.00μF for C2, 130V for ΔV2 in equation (VI) to find Q2.

    Q2=(6.00μF×106F1μF)(130V)=778×106C=778μC

Substitute 18.5mA for I, 4.00kΩ for R1, 7.00kΩ in equation (IX) to find V.

    V=(18.5mA×103A1mA)[(4.00kΩ×103Ω1kΩ)+(7.00kΩ×103Ω1kΩ)]=204V

Substitute 6.00μF for C2, 204V for V in equation (X) to find Q.

    Q=(6.00μF×106F1μF)(204V)=1222×106C=1222μC

Substitute 1222μC for Q, 778μC for Q2 in equation (XI) to find ΔQ.

    ΔQ=(1222μC×106C1μC)(778μC×106C1μC)=444×106C=444μC

Therefore, the amount of charge on the capacitor C2 is changed is 444μC_.

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Chapter 21 Solutions

Principles of Physics: A Calculus-Based Text

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