ESE 211 TEST 1
TEST 1A
ID
P1
P2
P3
Total
SECTION 1
109642383
105053217
110214593
2.5
2.5
3
108992184
8
3
3
109051837
7.5
3.5
8
109235299
108992555
5
2.5
2
108954569
0.5
3
4
109070054
110522580
1.5
2.5
4
110113799
109341280
5
2
4
109405205
109148009
129381048
3.5
1
2
109283241
109023986
4
0.5
1
109401324
110044365
109284082
109112314
3.5
0.5
1
110215741
5
1.5
3
110136161
SECTION 2
109547158
4.5
3
5
109464606
1.5
2.5
8
110042172
110049625
109044724
3.5
1.5
2
110655134
4.5
1.5
2
109369815
110195544
109386247
109136882
5
2.5
7
108691151
109063762
0.5
2.5
6
106662508
Averaged
P1
P2
TEST 1B
P3
Total
3
1.5
2.5
2.5
3
4
8.5
8
2
3
3
8
1
2.5
3
6.5
4.5
1.5
1
7
7
1
2
2.5
3
6
12
9.5
1.5
1.5
1
4
5
0.5
5
2
2
3
4
2
0
11
4.5
8
4
3
2
9
8
14
19
9.5
7.5
8
11
6.5
5.5
5
9.5
12.5
12
7
8
8
2
5
2
2.5
3.5
1
3
1
11
7.5
9.5
0.5
1.5
3
5
2.5
0
0
2.5
14.5
9
8.5
8
8
14
19
8
9.5
7.5
6.5
8
7
11
12
9.5
6.5
4
5.5
11
4.5
8
5
9.5
9
12.5
12
Missed
Missed
7
8
11
7.5
9.5
14.5
5
9
2.5
8.764706
14
12
# of students
10
8
6
4
2
0
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Score
Problem 1A (8 points):
For the given circuit below, R1=R3=1 KΩ，R2=2 KΩ. Vin=7 V, Iin1= 10mA，
Iin2=12 mA. Please find the current through R1, R2 and power consumed by resistor
R1.
R1
I_R2
Vin
R3
R2
Iin1
Iin2
Solution:
According to KCL, the relation of current I1 through R1 and current I2 though R2
could be expressed by:
I1=I2+10mA-12mA
(1)
According to KVL, the voltage drop on R1 and R2 together is 7V, therefore,
R1*I1+R2*I2=7V
(2)
By solving Eq. (1) & (2) I2 is found to be 3mA
From Eq. (1) current I1=1 mA. So. P=I2*R1=1 mW.
Problem 2A (4 points):
(1). Please derive the impedance of capacitor C, when vc=v0exp(jωt) is applied on
capacitor, where ω is the radian frequency. (hint v=Z*i)
(2). Please find the frequency if Vout/Vin=-6 dB, according to the circuit shown
below. Given that R=1 kΩ, C=100 nF
R


C
Vin
Vout


Solution:
The current through capacitor has the form
𝑑𝑣
i=C∗
𝑑𝑡
i = C ∗ jω ∗ v
v
So, the impedance of capacitance Z = = 1/jωC
i
(2).
The magnitude transfer function is T = 1/sqrt(1 + 𝜔2 𝜏 2 )
τ = RC = 1 ∗ 10−4 𝑠
If Vout/Vin=-6 dB, Vout/Vin=0.5
So, ω =
√3
𝜏
= 1.73 ∗ 104
𝑟𝑎𝑑
𝑠
, 𝑓=
𝜔
2𝜋
= 2.75 ∗ 103 𝐻𝑧
Problem 3A (8 points):
An oscilloscope measures the input and output voltage waveforms of a series RC
circuit (figure below). The vertical scale of the oscillogram is 0.2 V/div, and the
horizontal scale is 0.1 ms/div.
(1) What is the phase angle of the output voltage relative to the input voltage?
(2) Is this circuit operating in (a) ω«ω0 (b) ω≈ω0 or (c) ω»ω0? Give two reasons to
support your answer.
(3) You have 1kΩ resistor in the circuit. Find out the value of the capacitance and
draw the circuit with labels of the output and the ground.
Solution:
(1) Input is the one with 2Vp-p and the output is the one with 1.4Vp-p
Phase angle difference is
𝜙 = 360° ×
1
= 45°
8
Output is leading by 45°
(2) (b) ω=ω0
Phase angle difference is 45°
Magnitude of output voltage is 1⁄√2 of the input voltage.
(3)
𝑇 = 100𝜇𝑠 × 8 = 800𝜇𝑠
𝑓 = 𝑓0 =
1
= 1250 𝐻𝑧
800𝜇𝑠
𝜔0 = 2𝜋𝑓0 = 7850 𝑟𝑎𝑑/𝑠𝑒𝑐
𝜔0 =
1
𝑅𝐶
𝐶 = 127 𝑛𝐹
Problem 1B (8 points):
For the given circuit below, R1=2 KΩ，R2=2 KΩ, R3=3 KΩ. Vin=1 V, Iin1=
10mA，Iin2=12 mA. Please find the current through R1, R3, and power consumed
by resistor R1.
R1
R3
R2
Vin
I_R3
Iin1
Iin2
Solution:
According to KCL, the relation of current I1 through R1 and current I3 though R2
could be expressed by:
I1=I2+10mA-12mA
(1)
According to KVL, the voltage drop on R1 and R3 together is 1V,therefore,
R1*I1+R3*I2=1V
(2)
By solving Eq. (1) & (2) I2 is found to be 1mA
From Eq. (1) current I1=-1 mA. So. P=I2*R1=2 mW.
Problem 2B (4 points):
(1). Please derive the impedance of capacitor C, when vc=v0exp(jωt) is applied on
capacitor, where ω is the radian frequency. (hint v=Z*i)
(2). Please find the frequency if Vout/Vin=-12 dB, according to the circuit shown
below. Given that R=2 kΩ, C=200 nF
C


Vin
Vout
R


Solution:
The current through capacitor has the form
𝑑𝑣
i=C∗
𝑑𝑡
i = C ∗ jω ∗ v
v
So, the impedance of capacitance Z = = 1/jωC
i
(2).
The magnitude transfer function is T = 1/sqrt(1 + 𝜔2 𝜏 2 )
τ = RC = 1 ∗ 10−4 𝑠
If Vout/Vin=-6 dB, Vout/Vin=0.5
So, ω =
√3
𝜏
= 1.73 ∗ 104
𝑟𝑎𝑑
𝑠
, 𝑓=
𝜔
2𝜋
= 2.75 ∗ 103 𝐻𝑧
Problem 3B (8 points):
An oscilloscope measures the input and output voltage waveforms of a series RC
circuit (figure below). The vertical scale of the oscillogram is 0.2 V/div, and the
horizontal scale is 1 ms/div.
(1) What is the phase angle of the output voltage relative to the input voltage?
(2) Is this circuit operating in (a) ω«ω0 (b) ω=ω0 or (c) ω»ω0? Give two reasons to
support your answer.
(3) You have 5kΩ resistor in the circuit. Find out the value of the capacitance and
draw the circuit with labels of the output and the ground.
Solution:
(1) Input is the one with 2Vp-p and the output is the one with 1.4Vp-p
Phase angle difference is
ϕ=360°×1/8=45°
Output is lagging by 45°
(2) (b) ω=ω0
Phase angle difference is 45°
Magnitude of output voltage is 1⁄√2 of the input voltage.
(3)
T=1ms×8=8ms
f=f_0=1/8ms=125 Hz
ω_0=2πf_0=785 rad/sec
ω_0=1/RC
C=255 nF.