Download RLC circuit

RLC circuit
Submitted by: I.D. 066072570
The problem:
For the given RLC circle:
1. find the law of connecting inductances in series and in parallel in general.
What is the total induction of the given circle?
2. The switch is on the right hand side for 3τsec and after is moved to the left side.
• Find the resonance frequency of the system.
• find the current through the resistor as a function of time.
The solution:
a. inductance in series:
I1 = I2 ⇒ ε = (L1 + L2 )I˙1 = Lef f I˙ ⇒ L1 + L2 = Lef f
(1)
inductance in parallel:
I1 + I2 =
ε
ε
ε(L1 + L2 )
ε
L1 L2
1
1
1
+
=
⇒
=
= Lef f ⇒
=
+
L1 L2
L1 L2
I1 + I2
L1 + L2
Lef f
L1 L2
(2)
in our case:
Lef f =
(L1 + L2 ) · L3
L1 + L2 + L3
(3)
after 3τ (RC) we get:
−t
V = ε(1 − e Rc ) = ε(1 − e−3 ) ≈ 4.75v
(4)
and the equation of the circuit is:
q
q
+ IR + LI˙ = 0 ⇒ + Rq̇ + Lq̈ = 0
c
c
(5)
we shall start analysis with the simplest case- R=0, in this case we have:
q̈ = −
1
q
Lc
(6)
1
and the solution is:
q = Aeiω0 t
(7)
where
s
ω0 =
2
1
= 8 · 106 Hrz
Lef f c
3
(8)
substitute q = Aeiωt in the equation we get:
r
iR
Γ2
i
2
2
−ω +
ω + ω0 = 0 ⇒ ω = Γ ± ω02 −
L
2
4
(9)
where
Γ=
R
L
(10)
the solution is
i
h
0 i
q = Re Aei(ω + 2 Γ)t
(11)
where
r
0
ω =
ω02 −
Γ2
4
(12)
substitute
q(t = 0) = c · v(3τ )
(13)
we get
A = 4.75 · 10−9
(14)
to find the curent:
Γ
i
Γ
0 i
I = q̇ = RE i(ω 0 + Γ)Aei(ω + 2 Γ)t = −Ae− 2 t · ( cos(ω 0 t) + ω 0 sin(ω 0 t))
2
2
eventually substitute the given parameters we get:
2
2
2
I = −4.75 · 10−9 e−37509t 37509 · cos(8 · 106 t) + 8 · 106 sin(8 · 106 t)
3
3
3
Amper
(15)
(16)
The first plot presents a voltage on the resistor R and on the second one the current through it as
a function of time.
2
3