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

We differentiate Q(t) to get the current (I). Q(t) = [2 cos(30t) +  sin(30t)]  = I = -20 cos(30t) – 60 sin(30t) –  sin(30t) + 20 cos(30t)

I =  sin(30t)