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, = 2 coulombs, = 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).

L + RI + 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 = + I() d

Rewritten Equation:

L + R + = E(t)

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

0.1 + 2 + Q = 0

0.1 + 2 + 100 Q = 0

Characteristic Equation:

0.1 + 2r + 100 = 0

Use the quadratic formula to find roots.

r =

=

=

r =

= , =

Roots are complex so general solution is:

y=[cos() + sin()]

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

so;

Q(t) = [cos(30t)+sin(30t)]

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

Q(0)=[cos(30 0)+sin(30 0)]

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

2 = 1[1+0]

2 =

=0 amperes : ==0 we use this to solve for constant and t=0

Q(t) = [cos(30t)+sin(30t)]

Q(t) = ([cos(30t)+sin(30t)])

=[-10 cos(30t) – 30 sin(30t)] + [-10 sin(30t) + 30 cos(30t)]

0 = [-10 2 cos(30) – 30 2 sin(30 0)] + [-10 sin(30 0) + 30 cos(30 0)]

0 = -20 + 0 + 0 + 30

20 = 30

=

Therefore:

Q(t) = 2cos(30t) + sin(30t)

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

Q(t) = [2cos(30t) + sin(30t)]

= I = -20cos(30t) – 60sin(30t) – sin(30t) + 20cos(30t)

I = sin(30t)