Question #317938

For an electric circuit with L=0.05 henry, R=20 ohms and C=100*10^-6 farad, the applied emf is 100 volts. Prove that the charge q at time 't' is given by q(t) =0.01-e^(-200t) [0.01 cos(400t) +0.02 sin(400t) ] if initially q=0 and i=0.

1
Expert's answer
2022-03-28T10:59:10-0400

Solution.

Due to Kirchhoff's low:

UL+UR+UC=EU_L+U_R+U_C=E (1)

where

UL=Ldidt=Ld2qdt2U_L=L\frac{di}{dt}=L\frac{d^2q}{dt^2} ; UR=iR=dqdtRU_R=iR=\frac{dq}{dt}R ; UC=1CqU_C=\frac1Cq .

Ld2qdt2+Rdqdt+1Cq=EL\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac 1C q=E

d2qdt2+RLdqdt+1LCq=EL\frac{d^2q}{dt^2}+\frac RL\frac{dq}{dt}+\frac {1}{LC} q=\frac EL

RL=200.05=400\frac RL=\frac {20}{0.05}=4001LC=10.05100106=2105\frac {1}{LC}=\frac {1}{0.05\cdot 100\cdot 10^{-6}}=2\cdot10^5; EL=1000.05=2000\frac EL=\frac{100}{0.05}=2000

d2qdt2+400dqdt+2105q=2000\frac{d^2q}{dt^2}+400\frac{dq}{dt}+2\cdot10^5q=2000 (2)

dqdt=\frac{dq}{dt}= ddt(0.01e200t[0.01cos(400t)+0.02sin(400t)])\frac{d}{dt}(0.01-e^{-200t} [0.01 \cos(400t) +0.02 \sin(400t) ]) =e200t(6cos(400t)+8sin(400t))=e^{-200 t} (-6 \cos(400 t) + 8 \sin(400 t));

d2qdt2=400e200t(11cos(400t)+2sin(400t))\frac{d^2q}{dt^2}=400 e^{-200 t} (11 \cos(400 t) + 2 \sin(400 t)).

Substitution all these into (2) gives:

400e200t(11cos(400t)+2sin(400t))+400 e^{-200 t} (11 \cos(400 t) + 2 \sin(400 t))+ 400e200t(6cos(400t)+8sin(400t))+2105400\cdot e^{-200 t} (-6 \cos(400 t) + 8 \sin(400 t))+2\cdot 10^5 \cdot (0.01e200t[0.01cos(400t)+0.02sin(400t)])=(0.01-e^{-200t} [0.01 \cos(400t) +0.02 \sin(400t) ])= 20002000

We see that the left side is equal to the right one, which means q(t)q(t) satisfies Kirchhoff's law.


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