Section 6.3, Question 3

Find the current (I) in the RLC circuit, assuming that E(t) = 0 for t>0.

R = 2 ohms, L = 0.1 henrys, C = 0.01 farads,Q_o = 2 coulombs, I_o = 0 amperes

According to Kirchhoff’s Law the sum of the voltage drops in a closed RLC circuit is equal to the impressed voltage E(t).

LI' + RI + \frac{1}{C}Q = E(t)

This equation can be rewritten by replacing I with Q’ because charge stored in a capacitor is related to the current according to the equation below:

Q = Q_o + \int_{0}^{t} I(\tau) d\tau

Rewritten Equation:

LQ'' + RQ' + \frac{1}{C} = E(t)

Substituting the values of L, R, C, and E(t) we get:

0.1 Q'' + 2 Q' + \frac{1}{0.01} Q = 0

0.1 Q'' + 2 Q' + 100 Q = 0

Characteristic Equation:

0.1 r^2 + 2r + 100 = 0

Use the quadratic formula to find roots.

r = \frac{(-2)\pm\sqrt{2^2-4(0.1)(100)}}{2(0.1)}

= \frac{-2\pm\sqrt{-36}}{0.2}

= \frac{-2\pm6i}{0.2}

r = -10\pm30i

r_1 = -10 + 30i, r_2 = -10-30i

Roots are complex so general solution is:

y=e^{\lambda x}[C_1cos(\omega x) + C_2sin(\omega x)]

\lambda = (-10), \omega = (30) based on the roots we found

so;

Q(t) = e^{-10t}[C_1cos(30t)+C_2sin(30t)]

with Q_o=Q(0)=2 coulombs when t=0 we can solve for constant C_1

Q(0)=e^{-10(0)}[C_1cos(30 \times 0)+C_2sin(30 \times 0)]

cos(0)=1, sin(0)=0

2 = 1[C_1\times1+0]

2 = C_1

I_o=0 amperes : I_o=Q'_o=0 we use this to solve for constant C_2 and t=0

Q(t) = e^{-10t}[C_1cos(30t)+C_2sin(30t)]

\frac{d}{dt}Q(t) = (e^{-10t}[C_1cos(30t)+C_2sin(30t)]) \frac{d}{dt}

Q'(t)=[-10C_1 e^{-10t} cos(30t) – 30C_1 e^{-10t} sin(30t)] + [-10C_2 e^{-10t} sin(30t) + 30C_2 e^{-10t} cos(30t)]

0 = [-10\times 2 e^{-10(0)} cos(30\times 0) – 30 \times 2 e^{-10(0)} sin(30\times 0)] + [-10C_2 e^{-10(0)} sin(30\times 0) + 30C_2 e^{-10(0)} cos(30\times 0)]

0 = -20 + 0 + 0 + 30C_2

20 = 30C_2

\frac{2}{3} = C_2

Therefore:

Q(t) = 2e^{-10t}cos(30t) + \frac{2}{3} e^{-10t}sin(30t)

We differentiate Q(t) to get the current (I).

\frac{d}{dt}Q(t) = [2e^{-10t}cos(30t) + \frac{2}{3} e^{-10t}sin(30t)]\frac{d}{dt}

Q'(t) = I = -20e^{-10t}cos(30t) – 60e^{-10t}sin(30t) – \frac{20}{3} e^{-10t}sin(30t) + 20e^{-10t}cos(30t)

I = \frac{-200}{3} e^{-10t}sin(30t)