Answer to Question #317938 in Differential Equations for Sharlene

Question

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.

Answer 1

Solution.

Due to Kirchhoff's low:

$U_L+U_R+U_C=E$ (1)

where

$U_L=L\frac{di}{dt}=L\frac{d^2q}{dt^2}$ ; $U_R=iR=\frac{dq}{dt}R$ ; $U_C=\frac1Cq$ .

$L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac 1C q=E$

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

$\frac RL=\frac {20}{0.05}=400$; $\frac {1}{LC}=\frac {1}{0.05\cdot 100\cdot 10^{-6}}=2\cdot10^5$; $\frac EL=\frac{100}{0.05}=2000$

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

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

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

Substitution all these into (2) gives:

$400 e^{-200 t} (11 \cos(400 t) + 2 \sin(400 t))+$ $400\cdot e^{-200 t} (-6 \cos(400 t) + 8 \sin(400 t))+2\cdot 10^5 \cdot$ $(0.01-e^{-200t} [0.01 \cos(400t) +0.02 \sin(400t) ])=$ $2000$

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