1. In the circuit below R = 2 kQ, L = 250 mH, and C = 10 nF. The initial current Io = -4 A and the initial voltage on the capacitor is 0 V. Find the following: a. The damped form of the solution b. iR(0+) c. ic (0*) dv(0+) PP d. ic + iR dt e. v(t) for t≥0 C Vo L Io R V - 1. a. Overdamped (a = 25,000 rad/s, wo = 20,000 rad/s) b. iR (0+) = 0 A c. ic(0+) = 4A dv(0+) d. = 4 × 108 V/s dt e. v(t) = 13,333(e−10,000t. - e-40,000+)V
1. In the circuit below R = 2 kQ, L = 250 mH, and C = 10 nF. The initial current Io = -4 A and the initial voltage on the capacitor is 0 V. Find the following: a. The damped form of the solution b. iR(0+) c. ic (0*) dv(0+) PP d. ic + iR dt e. v(t) for t≥0 C Vo L Io R V - 1. a. Overdamped (a = 25,000 rad/s, wo = 20,000 rad/s) b. iR (0+) = 0 A c. ic(0+) = 4A dv(0+) d. = 4 × 108 V/s dt e. v(t) = 13,333(e−10,000t. - e-40,000+)V
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Can you explain/show how to find a through e?
![1. In the circuit below R = 2 kQ, L = 250 mH, and C = 10 nF. The initial current Io = -4 A and
the initial voltage on the capacitor is 0 V. Find the following:
a. The damped form of the solution
b. iR(0+)
c. ic (0*)
dv(0+)
PP
d.
ic
+
iR
dt
e. v(t) for t≥0
C
Vo
L
Io
R
V
-
1.
a. Overdamped (a = 25,000 rad/s, wo = 20,000 rad/s)
b. iR (0+) = 0 A
c. ic(0+) = 4A
dv(0+)
d.
= 4 × 108 V/s
dt
e. v(t) = 13,333(e−10,000t. - e-40,000+)V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b5049-d8dd-402a-aa3b-0cfd97dc82be%2F7c0ba44f-0712-42f3-9aab-0a7b8b4df091%2Fsg0w1o6_processed.png&w=3840&q=75)
Transcribed Image Text:1. In the circuit below R = 2 kQ, L = 250 mH, and C = 10 nF. The initial current Io = -4 A and
the initial voltage on the capacitor is 0 V. Find the following:
a. The damped form of the solution
b. iR(0+)
c. ic (0*)
dv(0+)
PP
d.
ic
+
iR
dt
e. v(t) for t≥0
C
Vo
L
Io
R
V
-
1.
a. Overdamped (a = 25,000 rad/s, wo = 20,000 rad/s)
b. iR (0+) = 0 A
c. ic(0+) = 4A
dv(0+)
d.
= 4 × 108 V/s
dt
e. v(t) = 13,333(e−10,000t. - e-40,000+)V
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