The example 2 on page 297 solves a system of equations for an electrical network. Analyze it and describe the steps taken for the solution (please write it in the computer step by step, like "step 1-...") you only have to write the steps, don't solve it. Book: Differential equations with boundary problems (7th edition)Topic: Laplace Transform EXAMPLE 2An Electrical NetworkSolve the system in (5) under the conditions E(t) = 60 V, L = 1 h, R = 50 N,C = 10-4 f, and the currents ij and iz are initially zero.SOLUTIONWe must solvedi+ 50i,dt6050(10-4) diz+ iz – i = 0dtsubject to i1(0) = 0, i2(0) = 0.Applying the Laplace transform to each equation of the system and simplifyinggives%3|60sl,(s) +501,(s)S-2001 (s) + (s + 200)I2(s)0,where I,(s)decomposing the results into partial fractions givesL{i;(t)} and I,(s)L{i>(t)}. Solving the system for I1 and I2 and60s + 12,0006/56/5601(s)||s(s + 100)?s + 100(s + 100)?S12,0006/56/5120,(s)s(s + 100)?(s + 100)2"Ss + 100Taking the inverse Laplace transform, we find the currents to bei,(t)6.-100te60te-100t|6i>(1)-100tе120te>-100t5 Note that both i(t) and i2(t) in Example 2 tend toward the valueE/R =as t→0. Furthermore, since the current through the capacitor isi3(t) = i(t) – i2(t) = 60te¬100r, we observe that i3(t) → 0 as t→ o.

Given system is di1dt+50i2=605010-4di2dt+i2-i1=0 (i) Subject to i10=0i20=0

Answered: EXAMPLE 2 An Electrical Network Solve…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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The example 2 on page 297 solves a system of equations for an electrical network. Analyze it and describe the steps taken for the solution (please write it in the computer step by step, like "step 1-...") you only have to write the steps, don't solve it.

Book: Differential equations with boundary problems (7th edition)

Topic: Laplace Transform

EXAMPLE 2
An Electrical Network
Solve the system in (5) under the conditions E(t) = 60 V, L = 1 h, R = 50 N,
C = 10-4 f, and the currents ij and iz are initially zero.
SOLUTION
We must solve
di
+ 50i,
dt
60
50(10-4) diz
+ iz – i = 0
dt
subject to i1(0) = 0, i2(0) = 0.
Applying the Laplace transform to each equation of the system and simplifying
gives
%3|
60
sl,(s) +
501,(s)
S
-2001 (s) + (s + 200)I2(s)
0,
where I,(s)
decomposing the results into partial fractions gives
L{i;(t)} and I,(s)
L{i>(t)}. Solving the system for I1 and I2 and
60s + 12,000
6/5
6/5
60
1(s)
||
s(s + 100)?
s + 100
(s + 100)?
S
12,000
6/5
6/5
120
,(s)
s(s + 100)?
(s + 100)2"
S
s + 100
Taking the inverse Laplace transform, we find the currents to be
i,(t)
6.
-100t
e
60te-100t
|
6
i>(1)
-100t
е
120te
>-100t
5

Note that both i(t) and i2(t) in Example 2 tend toward the value
E/R =
as t→0. Furthermore, since the current through the capacitor is
i3(t) = i(t) – i2(t) = 60te¬100r, we observe that i3(t) → 0 as t→ o.

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