Another Sample ES 250 Second Midterm Exam
1. This circuit has two inputs, vs and is, and one output io. The
output is related to the inputs by the equation
io = a is + b vs
Given the following two facts:
The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V.
and
The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V.
Determine the following:
The values of the constants a and b are a = ____0.6_____ and
b = __0.02___ A/V.
The values of the resistances are R 1 = ____30____ Ω and R 2 = ____20___ Ω.
0.45 = a ( 0.25) + b (15 )
From the 1st fact:
Substituting gives 0.45 = ( 0.60 )( 0.25 ) + b (15 ) ⇒ b =
Next, consider the circuit:
a i s = i o1 = i o
v s =0
⎛ R1 ⎞
=⎜
i
⎜ R1 + R 2 ⎟⎟ s
⎝
⎠
so
0.60 =
R1
R1 + R 2
⇒ 2 R1 = 3 R 2
and
b v s = i o2 = i o
i s =0
=
vs
R1 + R 2
so
0.02 =
1
R1 + R 2
0.30
= 0.60
0.50
0.45 − ( 0.60 )( 0.25 )
0.30 = a ( 0.50 ) + b ( 0 ) ⇒ a =
From the 2nd fact:
⇒ R1 + R 2 =
1
= 50 Ω
0.02
Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω.
15
= 0.02
2. Fill in the blanks in the following statements:
When R = 9 Ω then v R = _____3____V.
When R = ____27____Ω then v R = 5.4 V.
When R = ____12____Ω then i R = 300 mA.
Reduce this circuit using source transformations and equivalent resistance:
9
⎛ R ⎞
Now v R = ⎜
so the questions can be easily answered.
⎟ 9 and i R =
R + 18
⎝ R + 18 ⎠
3. Determine the values of the node voltages va, vb, vc and vo:
va = ___2.75___ V, vb = __2.8125__ V,
vc = ___2.25___ V, and vo = ___2.50___ V.
Due to the properties of the ideal op amp, va = 2.75 V and
vc = 2.25 V. The node equation at node c is
vb − vc
10 × 10
3
=
vc
40 ×10
3
⇒ vb =
5
v c = 2.8125 V
4
The node equation at node c is
vo − va
40 × 10
3
=
va − vb
10 × 10 3
⇒ v o = 5 v a − 4 v b = 2.5 V
4.
The input to this circuit is the voltage, vs. The output is the voltage vo. The voltage v b is used to adjust the
relationship between the input and output. Determine values of R 4 and v b that cause the circuit input and output
have the relationship specified by the graph
v b = ____4_____ V and R 4 = ____55____ kΩ.
(a) Label the node voltages as shown. The
node equations are
vs − va
R1
+
vb − va
R2
=
va
R3
and
va
R5
=
vo − va
R4
⎛ R5 ⎞
⇒ va = ⎜
v
⎜ R 4 + R 5 ⎟⎟ o
⎝
⎠
Solving these equations gives
vs ⎛ 1
v b ⎛ R1 R 2 + R 2 R 3 + R1 R 3
R5 ⎞
vb
1
1 ⎞
=⎜ +
+
va −
=⎜
×
vo −
⎟
⎟
R1 ⎜⎝ R1 R 2 R 3 ⎟⎠
R 2 ⎜⎝
R1 R 2 R 3
R 4 + R 5 ⎟⎠
R2
⎛
R 2 R3
R 4 + R5 ⎞
R1 R 3
R 4 + R5
So
vo = ⎜
×
vs +
×
× vb
⎟
⎜
R 5 ⎠⎟
R1 R 2 + R 2 R 3 + R1 R 3
R5
⎝ R1 R 2 + R 2 R 3 + R1 R 3
When R1 = R 2 = R 3 = 10 kΩ :
So
(b) The equation of the straight line is
We require
⎛ R 4 + R5 ⎞
R 4 + R5
vo = ⎜
vs +
× vb
⎟
⎜ 3 R5 ⎟
3 R5
⎝
⎠
R 4 + R5
R 4 + R5
a=
× vb
and b =
3 R5
3 R5
5
vs + 5
4
R 4 + R5 5
=
3 R5
4
vo =
When R 5 = 20 kΩ then R 4 = 55 kΩ . Next we require
5=
R 4 + R5
3 R5
5
× vb = vb
4
vb = 4 V
i.e.
5. The input to this circuit is the voltage:
v ( t ) = 4 e−20 t V for t > 0
The output is the current: i ( t ) = −1.2 e−20 t − 1.5 A for t > 0
The initial condition is i L ( 0 ) = −3.5 A . Determine the values of the
resistance and inductance:
R = ___5___ Ω and L = ___0.1____ H.
Solution: Apply KCL at either node to get
i (t ) =
v (t )
v (t ) ⎡ 1 t
⎤
+ i L (t ) =
+ ⎢ ∫ v (τ ) dτ + i ( 0 ) ⎥
0
R
R ⎣L
⎦
That is
−1.2 e
−20 t
4 e−20 t 1 t −20τ
4 e−20 t
4
e−20 t − 1) − 3.5
− 1.5 =
+ ∫ 4 e dτ − 3.5 =
+
(
0
R
L
R
L ( −20 )
⎛ 4 1 ⎞ −20 t 1
=⎜ −
− 3.5
⎟e +
5L
⎝ R 5L ⎠
Equating coefficients gives
And
1
− 3.5 ⇒ L = 0.1 H
5L
4 1
4
1
4
−1.2 = −
= −
= −2 ⇒ R =5 Ω
R 5 L R 5 ( 0.1) R
−1.5 =
6. The initial inductor current is i(0) =
25 mA.
Determine the values of the inductor
current at 2, 3, 6 and 9 seconds:
i(2) = ___−15___ mA,
i(3) = ___−55____ mA,
i(6) = _____5____ mA,
i(9) = ____65____ mA.
i (t ) =
so i (1) = 0.025 A
i (t ) =
1 t
0 dt + 0.025 = 0.025
100 ∫ 0
0 < t <1
for
−4 ( t − 1)
1 t
−4 dτ + 0.025 =
∫
1
100
100
for
1< t < 3
so i(2) = −0.015 A, and i ( 3) = −0.055 A
i (t ) =
2 ( t − 3)
1 t
2 dτ − 0.055 =
− 0.055
∫
100 3
100
for
3<t < 9
so i ( 9 ) = 0.065 A
i (t ) =
1 t
0 dτ + 0.065 = 0.065
100 ∫ 9
for
7.
a. When C = 10 F then Ceq = ___25__F.
b. When C = ___3.2___F then Ceq = 8 F.
C eq = C +
C ×C
5
+C = C
C +C
2
t >9
8. This circuit has reached steady state before the switch opens at
time t = 0. Determine the values of i L ( t ) , v C ( t ) and v R ( t )
immediately before the switch opens:
i L ( 0 − ) =____1____A, v C ( 0 − ) =____20____V
and
v R ( 0 − ) =_____−5_____V
Determine the value of v R ( t ) immediately after the switch opens:
v R ( 0 + ) =_____−4_____V
Solution: Because
• This circuit has reached steady state before the switch
opens at time t = 0.
• The only source is a constant voltage source.
At t=0−, the capacitor acts like an open circuit and the inductor
acts like a short circuit. From the circuit
25
25
i1 ( 0 − ) =
=
= 1.25 A ,
4 + ( 20 || 80 ) 4 + 16
⎛ 80 ⎞
iL (0 −) = ⎜
⎟ i1 ( 0 − ) = 1 A , v C ( 0 − ) = 20 i L ( 0 − ) = 20 V
⎝ 20 + 80 ⎠
and
v R ( 0 − ) = −4 i1 ( 0 − ) = −5 V
The capacitor voltage and inductor current don’t change
instantaneously so
v C ( 0 + ) = v C ( 0 − ) = 20 V and i L ( 0 + ) = i L ( 0 − ) = 1 A
Apply KCL at the top node to see that
i1 ( 0 + ) = i L ( 0 + ) = 1 A
From Ohm’s law
v R ( 0 + ) = −4 i1 ( 0 + ) = −4 V
(Notice that the resistor voltage did change instantaneously.)
9. After time t = 0, a given circuit is represented
by this circuit diagram.
a. Suppose that the inductor current is
i ( t ) = 21.6 + 28.4 e−4t mA
for t ≥ 0
R1 = ___6____ Ω and R 3 = ___40____ Ω.
Determine the values of R1 and R 3 :
b. Suppose instead that R1 = 16 Ω , R 3 = 20 Ω, the initial condition is i(0) = 10 mA, and the inductor current is
i ( t ) = A − B e− at for t ≥ 0 . Determine the values of the constants A, B, and a:
A = ___28.8__ mA, B = __−18.8___ mA and a = ____5___ s.
Solution:
The inductor current is given by i ( t ) = i sc + ( i ( 0 ) − i sc ) e − at
a. Comparing this to the given equation gives 21.6 = i sc =
4=
Rt
2
for t ≥ 0 where a =
R1
R1 + 4
( 36 )
1
τ
=
Rt
⇒ R1 = 6 Ω and
⇒ R t = 8 Ω . Next 8 = R t = ( R1 + 4 ) || R 3 = 10 || R 3 ⇒ R 3 = 40 Ω .
10
16
= 5 s . also i sc =
( 36 ) = 28.8 mA . Then
16 + 4
τ 2
= 28.8 + (10 − 28.8 ) e −5t = 28.2 − 18.8 e −5t .
b. R t = (16 + 4 ) || 20 = 10 Ω so a =
i ( t ) = i sc + ( i ( 0 ) − i sc ) e − at
1
=
10. a) Determine the time constant, τ, and the
steady state capacitor voltage, v(∞), when the
switch is open:
τ = _____3____ s and v(∞) = ____24____ V
b) Determine the time constant, τ, and the steady
state capacitor voltage, v(∞), when the switch is
closed:
τ = __2.25__ s and v(∞) = ____12____ V
Solution:
a.) When the switch is open we have
L
.
After replacing series and parallel resistors by equivalent resistors, the part of the circuit connected to the
capacitor is a Thevenin equivalent circuit with R t = 33.33 Ω . The time constant is
τ = R t C = 33.33 ( 0.090 ) = 3 s .
Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state.
Consequently, there is zero current in the 33.33 Ω resistor and KVL gives v(∞) = 24 V.
b.) When the switch is closed we have
This circuit can be redrawn as
Now we find the Thevenin equivalent of the part of the circuit connected to the capacitor:
So R t = 25 Ω and
τ = R t C = 25 ( 0.090 ) = 2.25 s
Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state.
Consequently, there is zero current in the 25 Ω resistor and KVL gives v(∞) = 12 V.