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) = [2
cos(30t) +
sin(30t)]
= I = -20
cos(30t) – 60
sin(30t) –
sin(30t) + 20
cos(30t)
I =
sin(30t)
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