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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)
