Linear Second-Order EquationsFind the current in the RLC circuit, assuming that E(t) = 0 for t > 0R = 4 ohms; L = .05 henrys; C = .008 farads; Q0 = -1 coulombs; I0 = 2 amperes We havePulting allLet Ithe differential equation,=the valuesd²f0.05dt² ++80.80 det intewe4dt2be the trialHence general sollution,havedet s+dtAextHence auxiliary equation,-> x =+2500 1 = 0I(t) =sollution ofI0.008x =eL-802-40td²Idt²c9² +80x + 250080diR = + = = 0곧dt2+ R=04x2500- 40 ± 30{ c₁cos (30t) + C₂ sin (30+)}[c1, c₂are constants ]

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Find the current in the RLC circuit, assuming that E(t) = 0 for t > 0 R = 4 ohms; L = .05 henrys; C = .008 farads; Q0 = -1 coulombs; I0 = 2 amperes

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Linear Second-Order Equations

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

R = 4 ohms; L = .05 henrys; C = .008 farads; Q₀ = -1 coulombs; I₀ = 2 amperes

We have
Pulting all
Let I
the differential equation,
=
the values
d²f
0.05
dt² +
+80.
80 det inte
we
4
dt2
be the trial
Hence general sollution,
have
det s
+
dt
Aext
Hence auxiliary equation,
-> x =
+2500 1 = 0
I(t) =
sollution of
I
0.008
x =
e
L
-802
-40t
d²I
dt²
c
9² +80x + 2500
80
di
R = + = = 0
곧
dt
2
+ R
=0
4x2500
- 40 ± 30
{ c₁cos (30t) + C₂ sin (30+)}
[c1, c₂
are constants ]

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