the characteristic equation are $12 = -a t Va – wi = -2, -50 and the corresponding response is v(1) = Aje + Ar We now apply the initial conditions to get A, and Az. 50 (8.5.1) v(0) = 5 = A, + Az (8.52) 5 + 0 1.923 x 10 x 10 dv(0) v(0) + Ri(0) -260 dt RC But differentiating Eq. (8.5.1). du 21 -24,e 50A2e dt
the characteristic equation are $12 = -a t Va – wi = -2, -50 and the corresponding response is v(1) = Aje + Ar We now apply the initial conditions to get A, and Az. 50 (8.5.1) v(0) = 5 = A, + Az (8.52) 5 + 0 1.923 x 10 x 10 dv(0) v(0) + Ri(0) -260 dt RC But differentiating Eq. (8.5.1). du 21 -24,e 50A2e dt
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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![Can you explain how
he made this step??!
Example 8.5
In the parallel circuit of Fig. 8.13, find v(t) for t > 0, assuming
v(0) = 5 V, i(0) = 0, L = I H, and C = 10 mF. Consider these cases:
R = 1.923 N, R = 5N, and R = 6.25 N.
Solution:
I CASE 1 If R = 1.923 ,
= 26
2RC
2 x 1.923 X 10 x 103
1
= 10
VLC
Vi x 10 x 10
Figure 8.13
A source-free parallel RLC circuit.
Since a > wo in this case, the response is overdamped. The roots of
the characteristic equation are
$12 = -a ± Va - wi = -2, -50
and the corresponding response is
U(1) = Aje1 + Ase S0
(8.5.1)
We now apply the initial conditions to get A, and A2.
(8.২)
v(0) = 5 = A, + A2
v(0) + Ri(0)
RC
đu(0)
5 +0
= -260
dt
1.923 x 10 x 10
But differentiating Eq. (8.5.1).
du
= -2Aje2 - 50Ase 5or
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73210801-93c3-466f-9c46-660a0b72a3a2%2F54a20a54-3b61-4b4a-bb5d-10bfe88f3e25%2Foba6m8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Can you explain how
he made this step??!
Example 8.5
In the parallel circuit of Fig. 8.13, find v(t) for t > 0, assuming
v(0) = 5 V, i(0) = 0, L = I H, and C = 10 mF. Consider these cases:
R = 1.923 N, R = 5N, and R = 6.25 N.
Solution:
I CASE 1 If R = 1.923 ,
= 26
2RC
2 x 1.923 X 10 x 103
1
= 10
VLC
Vi x 10 x 10
Figure 8.13
A source-free parallel RLC circuit.
Since a > wo in this case, the response is overdamped. The roots of
the characteristic equation are
$12 = -a ± Va - wi = -2, -50
and the corresponding response is
U(1) = Aje1 + Ase S0
(8.5.1)
We now apply the initial conditions to get A, and A2.
(8.২)
v(0) = 5 = A, + A2
v(0) + Ri(0)
RC
đu(0)
5 +0
= -260
dt
1.923 x 10 x 10
But differentiating Eq. (8.5.1).
du
= -2Aje2 - 50Ase 5or
dt
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