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An L-R-C series circuit has L = 0.400 H, C = 7.00 mF, and R = 320 Ω. At t = 0 the current is zero and the initial charge on the capacitor is 2.18×10-4C. (a) What are the values of the constants A and f in Eq. (30.28)q=Ae-(R/2L)tcos(1LC-R24L2t+ϕ)? (b) How much time does it take for each complete current oscillation after the switch in thiscircuit is closed? (c) What is the charge on the capacitor after thefirst complete current oscillation?

Short Answer

Expert verified

A) A=QAndϕ=0

B) The time taken to complete current oscillation is 0.0142s

C) The charge on the capacitor is9.75×10-7C

Step by step solution

01

Concept of the charge over the capacitor

The charge over the capacitor is given by equation q=Ae-(R/2L)tcos(1LC-R24L2t+ϕ)at t =0 the charge is the maximum, that is,q(t-0)=Q which can be written asQ=Ae0cos(0+ϕ)=Acos(ϕ)

02

Calculate the time taken to complete current oscillation

In the equation Q=Ae0cos(0+ϕ)=Acos(ϕ), RHS is maximum only whencosϕ=1so,A=Qand ϕ=0,

The angular frequency is given as ϖ=1LC-R24L2.At the need of the oscillationϖt=2π So, 2π=1LC-R24L2t. Substitute the values in the equationrole="math" localid="1664178436894" t=2π1LC-R24L2

t=2π10.400×7.00-320240.4002=0.0142s

Therefore, the time taken to complete current oscillation is 0.0142s

03

Calculate the charge on the capacitor after the first complete current oscillation

At the end of the first oscillation, the cosine term is 1, sinceϕ=1and role="math" localid="1664178711109" ϖt=2π.So,q=Qe-R/2Lt

Now substitute with t =0.0142 s and the givens to get q

role="math" localid="1664178926937" q=2.18×10-4Ce-320Ω0.0142s0.400H=9.75×10-7C

Therefore, the charge on the capacitor is9.75×10-7C

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