1

EE 182

Midterm I
[12:00]


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Last:
First:
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signature:



3/15/2022

 Draw a box around the final answers otherwise you will NOT get any credits And move your
solutions to the given boxes.
 Your phone must be turned off and kept in your bag.
 One (8.5x11) cheat sheet is allowed.
 Calculator is ok.
 No cell phone for calculator
 Only pencil and eraser

2

1. [8 7 15 pts]
Assume that the op-amp is ideal amp.
a) Find output voltage, ov
b) Find the current flow over the 9kΩ with direction?


a)

b) 9 5 Aki mΩ = ↑




a) 129v−
b) 5 Am ↑
3

2. [3 pts each 15 pts]
First order circuit is shown with DC input Vs and it is called CR circuit. The figure shows that the switch
position is set (switch up) that way when 0t < and the switch will be flip-down at 0t ≥

a) Find voltage over the C and R when 0t <
b) Find voltage over C and R when 0t ≥
c) Find current through C and R when 0t ≥
d) Plot voltage over C at [ ]0, 0,& 0t t t< = >
e) Plot current over C at [ ]0, 0,& 0t t t< = >





a)
The voltage over the C is fully charged @ 0t <
( ) 12VCv t =
And the current flow over the R is none since the capacitor is fully charged. So there is no current
flow.
( ) 0VRv t =
b) Only discharge occurs when 0t ≥ . It simply is a natural response
a) ( ) VC sv t v= ( ) 0 VRv t =
b) ( ) sC
t
v vt e τ
−
= ⋅ ( ) sR
t
v v et τ
−
−= ⋅
c) ( ) sR
t
i v et
R
τ
−
−= ⋅ ( ) sC
t
i v et
R
τ
−
−= ⋅
d)
e)
4

( ) ( )
( ) ( ) ( )
( )
0
0
1
t
s
t
C
R
t
t
s
c
t
s
v t v t e
v t i R where
dC v t e R
dt
dC v e R
dt
C v e R C R
d
v e R C
i t C v t
dt
τ
τ
τ
τ
τ
τ
τ
+
+
−
−
−
−
−
= =
= ⋅
 
= ⋅
⋅
=   
 
 
= ⋅ ⋅  
 
= ⋅ ⋅ −
= ⋅
⋅ = ⋅ 
1
sv RC
⋅ ⋅ −
t
s
t
e
v e τ
τ
−
−

−

= ⋅
 


( ) ( )
1 1
t
s
R C
t
s
t t
s s
v
i t i t
d
R
C v e
dt
C v e C v
e
e
RC
τ
τ
τ τ
τ
−
−
−
−
=
 
= ⋅  
   
= ⋅ ⋅ − = ⋅ ⋅ −   
− ⋅
 
=






5

3. [5 pts each 20 pts]
The switch position is set that way when 0t < (flipped down) and the switch position will be flip-up
when 0t ≥
a) @ 0t ≥ , show the detail procedure that voltage across the capacitor and plot it.
b) @ 0t ≥ , show the detail procedure that how
you got the voltage over the resistor (no need
to plot)
c) @ 0t ≥ , show the detail procedure that how
you got the current through the resistor and
plot it.
d) @ 10t ms= , how much power is delivered to
the resistor
a) ( ) ( ) ( )( )3 138.887.21012 1 12 1tC tv v et v e −− − − − ==  
b) ( ) ( ) 37.2 1012C
t
v ei t −
−
⋅
    
=
c) ( ) 37.2 100.005
t
i t e −
−
⋅
   
=


d) ( ) 3 7310 . wp ms mt = =

a)
The general equation for the DC input is
( ) ( ) ( )0 1t ty t y t e y eτ τ− −+  ≥ = + ∞ −  
The voltage over the capacitor is fully discharged 0t < since it is disconnected from the source,
then
( ) ( ) ( )( )
37.210
2 00 1 .4 3 .0072
12 1
t
t
e
y t y e where RC k Fτ τ μ
−
−
−
 
−  
 
≥ = ∞ − = = Ω = 
 

=


6

b)
The voltage over the resistor is

( )
( )
( )
( ) ( ) ( ) ( )3 377.2 1 .0 2 103
1
13 12 2.47. 12 10 2
t
t t
Rv t i R
dC v t R
dt
dC y e R
dt
F v e k v e
τ
μ − −
−
⋅
−−
−
⋅
   
= ⋅
 
= ⋅  
  
= ⋅ ∞ − ⋅     
  
= ⋅ ⋅ Ω =   
⋅     

c)
The current through the resistor is controlled by this equation ( ) ( )di t C v t
dt
 
=  
( ) ( )
( ) ( )
( )
3
3
2
7.2 1
.
0
3
7 10
2
1
13 12 7.2 1
0.005
0.00 =
0
373w 3.73
t
t
tdi
w
e
t C y e
dt
F v e
p mi R
τ
μ
−
−
−
−
−
⋅
⋅
−

    

 
= ⋅ ∞ −     
  
= ⋅   
⋅   
=
= =


7



d)
The power delivered to the resistor at the given time is
( ) ( )
( ) ( )
( ) ( ) ( )
3
3
10 10
7.2 10
3
10 10
1 2
A
.4
13 12 2.47.2
0.0012A
1.2m
10
t
p t ms i t ms R
dC y e k
dt
F v e k
τ
μ
−
−
−
⋅
−
⋅
−
= = = ⋅
  
= ⋅ ∞ − ⋅ Ω     
    = ⋅ ⋅ Ω 
⋅   
=
=



8

4. [15 pts]
Plot voltage and power


9



10

5. [10 pts]

a) Find ( )v t for the case of 0 ,0 &t − + = ∞ 
b) Write ( )v t and plot it.
c) Find ( )0.1v t ms=












a) ( )0v − ( )v ∞ τ
( )v t+ =
b)
c)
+5V

11



( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
1 1 2 1 2
2 1 2
2 1
1 2 1 1 2
2 1 2
1
2
20
1
2
: 5 2 2 4 5 2 2
: 5 2 2 3
: 4 5 2 2
: 5 2 5
6 6 5
2 5 5
3 1.11
@
.
3
0
0 277
1.1 1
.31 3 3
1
3
R x x
x
v
L k i v i i i v i i
L i i
where i i i
L i i v i i
L i i
i
i
v
i
t i
t
i
−
<

⋅ − = Ω − Ω ⇔ Ω ⋅ − = Ω − Ω
= − Ω +
  
=   
Ω
+
= −
⋅ − − = − Ω
= − +
−     
=   
−    
⋅ =

=

=



( )
1 1
2 1 2 3
3 1 2 3
2 1 2 3 2 3 1
3 1 2 3 2
2 3 1
2
3
3 1
3 1
2 1
1
: 0.277
: 4 2 5 5
: 5 2 5 5
: 4 2 5 5 2
: 5 2 5 5 5 5 5 2
21 1
5 25 5
sin
x x
L i
L i i i i where
L i i i
L i i i i i i
L i i i i i i
i i
i i
Matrix is gular to working pr
i i i i
i i i
=
⋅ = − + +
= − + +
⋅ = − + + ↔ + = −
= − + + ↔ + = +
−    
=     +   
+ −

= + −
ecision





12

6. [7 6 2 pts]

a) Find ( )v t for the case of 0 ,0 &t − + = ∞ 
b) Write ( )v t and plot it.
c) Find ( )0.1v t ms=


















a) ( )0 11 .754 V7v t − = ⋅ = ( )0 0.175vv t + = ( ) 0.5385vv t∞ =
b)
( ) 0.3635 0.5385 0.01850
t
y t e τ τ
−
−≥ = + =
c) ( )
30.110
0.3635 0. .5385 70 1 0.1 7y vs et m τ
−
⋅
−
 
+= = − =   


13

a)
( ) ( )0 1@ 0 : 74 1.75t v t −− = ⋅ =

( )@ 0t + 5.25V
Using superposition property,
7
5.25
1 7 5.25
1 79 0.7
0.525
0. 17 0.52
1
75 0. 5
v
v
v v
v
v
v
v
v
v v v v vΩ
 
= = + 
== + −
= −
=


The value of ( )v t right after the flipping the switch is
( )
( ) ( )
0
3
0.17
1
5
0.5381 57
v t v
vv
t
t v
+
=
= ∞
=
=
∞ =



The equation for the voltage over the resistor is
( ) ( ) ( )( ) ( ) ( )
( )
0.3635 0.5385 0.0185
1
8
0.0183 50 0 || 0 8
0.175 0.53 5 0.5385
t
t
t
y t y y e y
e
RC mF
e
τ
τ
τ
τ
τ
−
+
−
−
≥ = −
−
− + =
∞ + ∞ = = ⋅ =
= +
=


c) ( )
30.110
0.3635 0. .5385 70 1 0.1 7y vs et m τ
−
⋅
−
 
+= = − =   