Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 97P

(a)

To determine

The time constant of the RC circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 97P

The time constant of the RC circuit is 8.7×104s_.

Explanation of Solution

Write the expression for the charge in the capacitor during charging.

Q(t)=Q0et/τ        (I)

Here, Q(t) is the charge in the capacitor during charging, Q0 is the charge in the capacitor at time t=0, τ is the time constant.

Use equation (I) to solve for τ.

τ=tln(Q(t)Q0)        (II)

Conclusion:

Substitute 4.0×103s for t, (0.010)Q0 for Q(t) in equation (II) to find τ.

    τ=4.0×103sln((0.010)Q0Q0)=4.0×103sln(0.010)=8.7×104s

Therefore, the time constant of the RC circuit is 8.7×104s_.

(b)

To determine

The resistance of the flashbulb in the circuit.

(b)

Expert Solution
Check Mark

Answer to Problem 97P

The resistance of the flashbulb in the circuit is 1.2Ω_.

Explanation of Solution

Write the expression for the time constant.

τ=RC        (III)

Here, R is the resistance, C is the capacitance.

Use equation (III) to solve for R.

R=τC        (IV)

Write the expression for the energy stored in the capacitor.

U=12C(ΔV)2        (V)

Here, U is the energy stored in the capacitor, ΔV is the voltage across the capacitor.

Use equation (V) to solve for C.

C=2U(ΔV)2        (VI)

Use equation (VI) and (II) in (IV) to solve for R.

R=tln(Q(t)Q0)2U(ΔV)2        (VII)

Conclusion:

Substitute 4.0×103s for t, (0.010)Q0 for Q(t), 32J for U, 300V for ΔV in equation (VII) to find R.

    R=4.0×103sln((0.010)Q0Q0)2(32J)(300V)2=(4.0×103s)(300V)22(32J)ln(0.010)=1.2Ω

Therefore, the resistance of the flashbulb in the circuit is 1.2Ω_.

(c)

To determine

The maximum power dissipated in the flashbulb.

(c)

Expert Solution
Check Mark

Answer to Problem 97P

The maximum power dissipated in the flashbulb is 74kW_.

Explanation of Solution

Write the expression for the maximum power dissipated.

Pmax=V2R        (VIII)

Here, Pmax is the maximum power dissipated.

Conclusion:

Substitute 300V for V, 1.2Ω for R in equation (VIII) to find Pmax.

    Pmax=(300V)21.2Ω=74×103W=74kW

Therefore, the maximum power dissipated in the flashbulb is 74kW_.

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Chapter 18 Solutions

Physics

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