Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below is di 3 dt q= so diz L dt di 2 -R₁ dt E(t) = Li₁' + L Ai₁ E diz -R1 dt +L Then 13 = q' = CR₁12 - CR₂13' + R₂ 00000 L diz di3 dt dt Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + +L + R₂ so that in terms of C, R₁, R₂, 12 and 13 we have the following. di 3 dt + di3 dt + R₁₂ = E(t) iz R₂ 2/3 R₁ + R₁/2 = E(t) + = 0. = 0. IC
Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below is di 3 dt q= so diz L dt di 2 -R₁ dt E(t) = Li₁' + L Ai₁ E diz -R1 dt +L Then 13 = q' = CR₁12 - CR₂13' + R₂ 00000 L diz di3 dt dt Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have E(t) = Li₁' + +L + R₂ so that in terms of C, R₁, R₂, 12 and 13 we have the following. di 3 dt + di3 dt + R₁₂ = E(t) iz R₂ 2/3 R₁ + R₁/2 = E(t) + = 0. = 0. IC
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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![Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below is
di 3
dt
9 =
so
E(t) = Li₁' +
L
diz
L
dt
di 2
-R₁ dt
diz
dt
diz
-R1 dt
Ai₁
E
+ L
Then
13 = 9' = CR₁12 - CR₂13'
+ R₂
00000
L
Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have
E(t) = Li₁' +
+L
+ R₂
so that in terms of C, R₁, R₂, 12 and 13 we have the following.
di3
dt
di 3
dt
+
di 3
dt
+ R₁₂ = E(t)
iz R₂
1/13
R₁
+ R₁/2 = E(t)
+
= 0.
= 0.
C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74a50780-bdf2-45e2-b018-f4cc84bd693f%2F8764ea7f-deda-48ff-b987-8e9761c26853%2F2akkmnk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Show that a system of differential equations that describes the currents i(t) and i3(t) in the electrical network shown in the figure below is
di 3
dt
9 =
so
E(t) = Li₁' +
L
diz
L
dt
di 2
-R₁ dt
diz
dt
diz
-R1 dt
Ai₁
E
+ L
Then
13 = 9' = CR₁12 - CR₂13'
+ R₂
00000
L
Using Kirchhoff's first law we have i₁ = 12 + 13. By Kirchhoff's second law, on each loop we have
E(t) = Li₁' +
+L
+ R₂
so that in terms of C, R₁, R₂, 12 and 13 we have the following.
di3
dt
di 3
dt
+
di 3
dt
+ R₁₂ = E(t)
iz R₂
1/13
R₁
+ R₁/2 = E(t)
+
= 0.
= 0.
C
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