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Solution of
ECE 300 Test 8 F09
1.
In the space provided graph this voltage versus time.
Put a scale on the vertical axis so that actual voltages
could be read from the graph.
v
t
( ) =
4u
−
t
( )
−
6u 2
−
t
( ) +
5u
t
−
4
( )
2.
Find the numerical values of the capacitor current and voltage and the switch current at the times indicated.
v
C
0
−
( ) =
____________V
v
C
0
+
( ) =
____________V
v
C
∞
( ) =
____________V
i
C
0
−
( ) =
____________mA
i
C
0
+
( ) =
____________mA
i
C
∞
( ) =
____________mA
i
s
0
−
( ) =
____________mA
i
s
0
+
( ) =
____________mA
i
s
∞
( ) =
____________mA
What is the numerical time constant of this circuit after the switch is closed.
τ
= ____________ s
I
1
=
2mA ,
R
2
=
5k
Ω
,
R
3
=
10k
Ω
,
V
4
=
15V ,
C
5
=
4
μ
F
The circuit has been in the configuration with the switch open for a very long time and all currents and voltages are
constant.
For the current source only:
The current source current must all flow through resistor
R
2
developing a voltage of 10V (positive on top).
Then
the capacitor voltage would also be 10V.
The switch current is zero.
For the voltage source only:
No current flows anywhere in the circuit so the capacitor voltage must be 15V.
Overall:
The capacitor voltage is 5V and the capacitor current is zero.
At time
t
=
0
+
:
10
10
10
10
v(
t
)
t
I
1
R
2
t
= 0
C
5
V
4
R
3
v
C
i
C
i
s
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View Full Document The capacitor voltage remains at 5V.
All the current source current now flows through the switch.
The voltage
across
R
3
is 10V (positive on the left).
Therefore the current through the capacitor, voltage source and
R
3
is 1mA (counter
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This note was uploaded on 05/01/2010 for the course ECE 300 taught by Professor Watson during the Spring '09 term at CarsonNewman.
 Spring '09
 Watson
 Volt

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