Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Practice Exam 3 on Circuit Analysis I - Fall 2005 | EGR 214, Exams of Electrical Circuit Analysis

Material Type: Exam; Class: Circuit Analysis I; Subject: Engineering; University: Grand Valley State University; Term: Fall 2005;

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

koofers-user-9xi-2
koofers-user-9xi-2 🇺🇸

10 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
EGR 214 – Fall 2005 E3 -1
Grand Valley State University
Prof. Bogdan Adamczyk
Exam 3
Name:
This is a closed book, closed notes exam. Calculators are allowed.
Provide complete solutions (not just the answers) to all problems. The final answers will
constitute only a part of your grade; the majority of it will be based on the process
(explanation) of how the answers have been obtained. This means: Explain in words what
you are doing and why while showing the calculations. The numerical results (with 4
digits after the decimal point) should include proper units.
Prior to the circuit has been in the steady-state condition. At 0=t0
=
t the switches
move from positions A to positions B.
B
A
2K 6K
+
t = 0
3K
B
+
v1(t) 4K
0.25
µ
F
+
v(t)
i(t)
2K
A
t = 0
1K
20 mA 60 V
Determine:
a)
(
)
0i
b)
(
)
0v
c)
(
)
+
0i
d)
(
)
+
0v
e)
()
i
f)
()
v
g)
()
0, >tti
h)
()
0, ttv
i)
()
0,
1>ttv
pf3
pf4

Partial preview of the text

Download Practice Exam 3 on Circuit Analysis I - Fall 2005 | EGR 214 and more Exams Electrical Circuit Analysis in PDF only on Docsity!

Grand Valley State University Prof. Bogdan Adamczyk Exam 3

Name:

This is a closed book, closed notes exam. Calculators are allowed.

Provide complete solutions (not just the answers) to all problems. The final answers will constitute only a part of your grade; the majority of it will be based on the process (explanation) of how the answers have been obtained. This means: Explain in words what you are doing and why while showing the calculations. The numerical results (with 4 digits after the decimal point) should include proper units.

Prior to t = 0 the circuit has been in the steady-state condition. At t = 0 the switches move from positions A to positions B.

B 2K^ A 6K

t = 0

3K

B

v 1 (t) 4K

0.25 μ F

v(t)

i(t)

2K

t = 0 A

1K

20 mA (^) 60 V

Determine:

a) i ( 0 −)

b) v ( 0 −)

c) i ( 0 +)

d) v ( 0 +)

e) i ( ∞)

f) v ( ∞)

g) i ( ) t , t > 0

h) v ( ) t , t ≥ 0

i) v 1 ( ) t , t > 0

Grand Valley State University Prof. Bogdan Adamczyk Exam 3

Name: Solution:

First, determine the initial voltage across the capacitor. Prior to t = 0 , the relevant circuit can be redrawn as

1K 2K

0.25 μ F

2K^ V^0

I

i(0-)

V 0

o pen circuit 20 mA 4K

Figure 2.

According to current divider

I ( 20 )( 10 ) 4. 4444 [ mA ]

V 0 = v ( 0 −^ ) = 2000 I + 4000 I = 26. 6667 [ V ] (2.1)

i ( 0 −^ ) = 0 [ mA ] (2.2)

After t = 0 the circuit becomes

2K 6K

0.25 μ F^ −

4K

v(t),V 0

i(t)

v 1 (t)

3K

P

60 V

Q

Figure 2.

Grand Valley State University Prof. Bogdan Adamczyk Exam 3

Name: Therefore,

i ( 0 +^ ) =− 4 [ mA ] (2.8)

and

i (∞ ) = 0 [ mA ] (2.9)

Now, return to Figure 2.3, redrawn here

2K 6K

0.25 μ F^ −

4K

v(t),V 0

i(t)

v 1 (t)

3K

P

60 V

Q

Figure 2.

KVL in the leftmost loop yields

− v ( ) t −( 2000 )( 0. 008 ) e −^1200 t^ + v 1 ( ) t = 0 (2.10)

or

v 1 ( ) t = 13. 3333 + 5. 3333 e −^1200 t^ [ V ], t > 0 (2.11)