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Principles Of Electric Circuits
10th Edition
ISBN: 9780134879482
Author: Floyd, Thomas L.
Publisher: Pearson,
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Students have asked these similar questions
A singl core cable of voltage 30 kv.
The diameter of Conductor is 3 cm.
The diameter of cable is 25 cm. This
cable has Two layer of insulator having
arelative permittivity 5-3 respectively
of
The ratio of
maximum electric stress
of
maximum electric stress
8
First layer to the
of second layer is 10 Find &
1- The thickness of each layers.
3-
The voltage of each
layers. §.
Layers
The saving in radius of cable if
another ungrading cable has the
Same maximum electric stress, Total
village, Conductor diameter of
grading cable.
66 KV sing care Cable has
a drameter of conductor of 3 cm.
The radius of cable is 10 cm.
This Cable house Two relative permmitivity
of insulation 6 and 4 respectively.
If The ratio of maximum electric stress
of first layer to the maximum eledric
streep & second layer is s
1- find the village & each layers.
2- Min- electric stress J Cable
3- Compare the voltage of ungrading
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relectric stresses.
Prelab Information
1. Laboratory Preliminary Discussion
First-order Low-pass RC Filter Analysis
The first-order low-pass RC filter shown in figure 1 below represents all voltages and currents in the time domain. It is of course
possible to solve for all circuit voltages using time domain differential equation techniques, but it is more efficient to convert the
circuit to its s-domain equivalent as shown in figure 2 and apply Laplace transform techniques.
vs(t)
i₁(t)
+
R₁
ww
V₁(t)
12(t)
Lic(t)
Vout(t)
=
V2(t)
R₂
Vc(t)
C
Vc(t)
VR2(t)
= V2(t)
+
Vs(s)
Figure 1: A first-order low-pass RC filter represented in the time domain.
I₁(s)
R1
W
+
V₁(s)
V₂(s)
12(s)
Ic(s)
+
Vout(S)
==
Vc(s)
Vc(s)
Zc(s)
=
=
VR2(S)
V2(s)
Figure 2: A first-order low-pass RC filter represented in the s-domain.
Chapter 8 Solutions
Principles Of Electric Circuits
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