ELEC 273 Part 12
Old Exam Questions
1.
2.
3.
4.
Old Exam Questions with Solutions
2016 Final Exam
2017 Final Exam
2018 Final Exam
Copyright © 2020 C.W. Trueman
1
Mesh Analysis
Write mesh equations.
Method:
1.Write a “constraint
equation” for the current
source.
2.Write a set of mesh
equations, including the
supermesh.
Solution
F
E
G
For the current generators:
B
𝑖 =1
C
H
𝑖 −𝑖 = 3
I
Mesh path ABCDA
4 𝑖 −𝑖 +3−2 𝑖 −𝑖
A
−8 𝑖 −𝑖
Supermesh DCBEFGHIJD
8 𝑖 −𝑖 +2 𝑖 −𝑖 −2 𝑖
=0
+ 4 − 6𝑖 − 10𝑖 − (−5) = 0
D
J
Node Analysis
Method:
1.Write “constraint equations” for
the voltage sources.
2.Write a KCL equation for the
supernode.
3.Write the “controlling current” 𝑖
in terms of the node voltages and
then eliminate 𝑖 from the node
equations.
Note that this circuit has a
supernode that includes two
voltage sources!
Solution
Constraint Equations
𝑣 − 2𝑖 = 𝑣
𝑣 − 12 = 𝑣
Controlling current
𝑣
𝑖=
4
Eliminate 𝑖 from 𝑣 − 2𝑖 = 𝑣 to
get
𝑣
𝑣 −2 =𝑣
4
Which simplifies to
𝑣
𝑣 − =𝑣
2
KCL Equation
KCL for the supernode:
flows BOTH into and out of the
supernode!
−
𝑣
𝑣
𝑣 −𝑣
𝑣 −𝑣
𝑣
− −
+
− =0
4
1
2
2
4
−
𝑣
𝑣
𝑣
− − =0
4
1
4
Hence the three equations are:
𝑣
𝑣 − =𝑣
2
𝑣 − 12 = 𝑣
𝑣
𝑣
𝑣
− − − =0
4
1
4
𝑣 −𝑣
2
Thevenin Equivalent Circuit
Is=2A
Vs=8V
𝑉 =8
𝐼 =2
Find the Thevenin Equivalent Circuit at terminals a-b.
Method:
1.Find the open-circuit voltage, 𝑉 , and then 𝑉 = 𝑉
To find 𝑅 :
Either
find the short-circuit current, 𝐼 and 𝑅 = ,
or
find the resistance of the “dead circuit”.
This is a “judgement call” based on your experience.
In general when the circuit has dependent sources, the
short-circuit current method is better.
Solution
𝑣
𝑣
𝑣
−8
𝑣
8
2
Is=2A
Vs=8V
+
𝑣
−
Open Circuit Voltage:
KCL at 𝑣
𝑣
𝑣 − 𝑣 −8
𝑣 −𝑣
− −
−
=0
1
2
2
KCL at the supernode:
𝑣 − 𝑣 −8
𝑣 −𝑣
𝑣
+2+
−
=0
2
2
2
Multiply 1st eqn by 2:
−2𝑣 − 𝑣 + 𝑣 − 8 − 𝑣 + 𝑣 = 0
−4𝑣 + 2𝑣 = 8
8 − 2𝑣
𝑣
𝑣 =
= −2 +
−4
2
nd
Multiply 2 eqn by 2:
𝑣 − 𝑣 −8 +4+𝑣 −𝑣 −𝑣 = 0
2𝑣 − 3𝑣 = −8 − 4
𝑣
2(−2 +
) − 3𝑣 = −12
2
−4 + 𝑣 − 3𝑣 = −12
−2𝑣 = −8
𝑣 =4
Solution
0 volts
𝑣
−8
0
8
2
Is=2A
Vs=8V
𝑖
0
0 volts
Short circuit current:
KCL at 𝑣 is almost the same as for the open-circuit test,
except that the voltage at terminal “a” is zero:
𝑣
𝑣 − −8
𝑣
− −
− =0
1
2
2
−2𝑣 − 𝑣 + −8 − 𝑣 = 0
−4𝑣 = 8
𝑣 = −2
KCL at the supernode:
𝑣 − −8
𝑣
+2+ −𝑖 =0
2
2
Multiply by 2
𝑣 − −8 + 4 + 𝑣 − 2𝑖 = 0
2𝑣 − 2𝑖 = −8 − 4
−2𝑖 = −12 − 2𝑣
𝑖 =6+𝑣
𝑖 = 6 + (−2)
𝑖 =4
Thevenin Equivalent Circuit
𝑣 = 4 volts
𝑖 = 4 amps
So
𝑣 =4
𝑣
𝑅 =
𝑖
Check: Dead circuit method
𝑅
𝑅
𝑅
𝑅
= ( 2 ∥2 +1 ∥2
= 1+1 ∥2
=2∥2
=1
=
4
=1Ω
4
Maximum Power Transfer Theorem
Find R such that it dissipates the maximum amount of power.
How much power does R dissipate?
Method:
1.Find the Thevenin Equivalent.
2.Choose 𝑅 = 𝑅
3.Power is 𝑃 =
Find the open-circuit voltage:
𝑖
+
𝑣
−
𝑖
Current in the 80+20 branch: 𝑖 =
= 0.4 amps
Current in the 10+90 branch: 𝑖 =
= 0.4 amps
𝑣 =voltage across the 10 ohm resistor + voltage across the 80 ohm
resistor
𝑣 = +10𝑖 − 80𝑖
𝑣 = 10𝑥0.4 − 80𝑥0.4 = 4 − 32 = −28 volts
Check:
𝑣 =voltage across the 90 ohm resistor + voltage across the 20 ohm
resistor
𝑣 = −90𝑖 + 20𝑖
𝑣 = −90𝑥0.4 + 20𝑥0.4 = 36 − 8 = −28 volts
Dead circuit method, easy for this circuit!
80 Ω
20 Ω
𝑅
10 Ω
𝑅 = 80 ∥ 20 + 10 ∥ 90 =
80𝑥20
10𝑥90
+
= 16 + 9 = 25 Ω
80 + 20 10 + 90
90 Ω
Maximum Power:
Find R such that it dissipates the maximum amount of power.
How much power does R dissipate?
𝑉 = −28 volts
A
𝑅 = 25 Ω
B
Find R such that it dissipates the maximum amount of power. How much power does R dissipate?
Method:
1.Find the Thevenin Equivalent – done!
2.Choose 𝑅 = 𝑅 : Hence choose the load as 𝑅 = 25 Ω
3.Power is 𝑃 =
: Hence the maximum power is 𝑃 =
(
)
= 7.84 watts.
Op Amp Circuit
The op-amps are ideal.
Find the output voltage 𝑣 .
Method:
The voltage across the input of the op-amp is
zero.
Use node analysis and write KCL equations.
Solution
𝑣
+0 −
0
𝑣
𝑣
The input resistance of the op-amp is high so 𝑖 = 0 so the voltage across 𝑅 is zero.
The voltage across the input to the op-amp is zero.
Hence the voltage at the “-” input equal to 𝑣 .
𝑅 𝑣 = (𝑅 + 𝑅 )𝑣
KCL at the top of 𝑅 :
𝑅 +𝑅
𝑣 −𝑣
𝑣
𝑣 =
𝑣
−
=0
𝑅
𝑅
𝑅
𝑅 𝑣 −𝑅 𝑣 −𝑅 𝑣 =0
Solution
0
+0 −
0
0
𝑣
𝑅 +𝑅
𝑣
𝑅
KCL at the input of the 2nd op amp:
𝑣
𝑣
+
=0
𝑅
𝑅
𝑅 𝑣 +𝑅 𝑣 =0
𝑣
=
𝑣
𝑣
𝑅
𝑣
𝑅
𝑅 𝑅 +𝑅
=−
𝑣
𝑅
𝑅
=−
First Order Transient
For 𝑡 < 0, the switch has been in position
1 for a long time. At 𝑡 = 0 the switch
changes to position 2 and stays there for
𝑡 > 0.
1.Find the initial condition 𝑣 0 .
2.Find the final condition 𝑣 (∞). This is the
“forced response”.
3.Find the time constant 𝜏.
4.Find the “natural response”.
5.Add up the forced response plus the
natural response and evaluate the arbitrary
constant using the initial condition.
1
2
For 𝑡 < 0, the switch has been in position
1 for a long time. At 𝑡 = 0 the switch
changes to position 2 and stays there for
𝑡 > 0.
1.Find the initial condition 𝑣 0 .
1
2
For 𝑡 < 0 the switch has been in position #1 for a long time. Find 𝑣 (0).
After the switch has been in position 1 “for a long time”, the voltage across the capacitor becomes constant
and the current 𝑖 = 𝐶
is zero. The capacitor behaves as an open circuit.
6Ω
6Ω
1
1
0.5
6Ω
0.1 F
+
𝑣 (0)
−
0.5
6Ω
0.1 𝜇F
+
𝑣 0 = −0.5x6 = −3
−
For 𝑡 < 0, the switch has been in position 1 for a long time. At 𝑡 = 0 the switch changes to position 2 and stays there for
𝑡 > 0.
1.Find the initial condition 𝑣 0 .
Just before the switch changes:
6Ω
1
0.5
+
𝑣 0 = −3
−
6Ω
Just after the switch changes.
So the initial condition is 𝑣 (0) = −3 volts
1
2
2Ω
6Ω
0.5
The stored energy in the capacitor is 𝑤 = 𝐶𝑣 .
The stored energy cannot change instantaneously
when the switch changes position.
So the voltage across the capacitor 𝑣 = −3 volts
just before the switch changes is the same as the
voltage across the capacitor 𝑣 = −3 volts just
after the switch changes.
6Ω
0.1F
+
𝑣 0 = −3
−
24 Ω
8Ω
1
64
2.Find the final condition 𝑣 (∞). The final value as 𝑡 → ∞ is the “forced response”.
As 𝑡 → ∞, the switch has been in position 2 for a long time, and
the voltage across the capacitor becomes constant and the
current 𝑖 = 𝐶
is zero. The capacitor behaves as an open
circuit.
2
24 Ω
2Ω
0.1 𝜇F
+
𝑣 ∞
−
8Ω
64
Write a node equation:
𝑣
𝑣 − 64
− −1−
=0
8
24
−3𝑣 − 24 − 𝑣 + 64 = 0
−4𝑣 = −40
𝑣 = 10
𝑣
2
+
𝑣 ∞
−
1
2Ω
8Ω
24 Ω
1
64
Since there is no current in the 2 Ω resistor,
𝑣 (∞) = 10
3.Find the time constant 𝜏.
2
24 Ω
2Ω
Use the ‘dead circuit’ method to find the resistance across the
capacitor terminals.
8 parallel 24 =
𝑅
8Ω
=
=6
𝑅 =2+6 = 8Ω
𝐶 = 0.1 F
The time constant is
𝑣
𝜏 = 𝑅 𝐶 = 8𝑥0.1 = 0.8 seconds
4.Find the “natural response”.
5.Add up the forced response plus the natural response and evaluate the arbitrary constant using the initial condition.
Solution
4.The “natural response” for a first-order circuit is 𝑣 𝑡 = 𝐴𝑒 .
5.The total response is the sum of the forced response 𝑣 (∞) = 10 plus the natural response 𝑣 𝑡 = 𝐴𝑒 .
𝑣 𝑡 = 10 + 𝐴𝑒 .
Where the time constant is 𝜏 = 0.8 seconds
The initial condition is 𝑣 (0) = −3 volts
𝑣 0 = 10 + 𝐴𝑒
= −3
10 + 𝐴 = −3
𝐴 = −13
The complete response is
𝑣 𝑡 = 10 − 13𝑒
Note that this question did not ask you to find the differential equation!
First Order Transient
The capacitor is uncharged at 𝑡 = 0.
Write a differential equation for 𝑣 𝑡 .
Solve the differential equation to find 𝑣 𝑡 .
Remark:
This question ask you to write a differential
equation and then solve it. So you cannot use
the “magic formula” to answer this question!
+𝑣 −
0.1 F
3A
2Ω
8Ω
3A
+𝑣 −
Write a differential equation for 𝑣 𝑡 .
We can make the circuit much simpler with “source
transformations”.
Replace 3 A in parallel with 2 ohms with 3x2=6 volts in series
with 2 ohms.
0.1 F
3A
2Ω
8Ω
3A
Replace 3 A in parallel with 8 ohms with 3x8=24 volts in series
with 8 ohms.
KVL:
6 − 2𝑖 − 𝑣 − 8𝑖 + 24 = 0
−10𝑖 − 𝑣 = −30
10𝑖 + 𝑣 = 30
𝑑𝑣
𝑖 = 0.1
𝑑𝑡
𝑑𝑣
10x0.1
+ 𝑣 = 30
𝑑𝑡
𝑑𝑣
+ 𝑣 = 30
𝑑𝑡
This is the differential equation!
+𝑣 −
2Ω
0.1 F
8Ω
24 v
6v
𝑖
Solve the differential equation to find 𝑣 𝑡 .
𝑑𝑣
+ 𝑣 = 30
𝑑𝑡
The D.E. has the form
+ 𝑣 = constant
So the time constant is 𝜏 = 1.
The forced response is always a constant with D.C. sources,
so use 𝑣 = 𝐵, where 𝐵 is a constant.
𝑑𝐵
+ 𝐵 = 30
𝑑𝑡
0 + 𝐵 = 30 so 𝐵 = 30
Forced response: 𝑣 = 30
Note that this is the “final value” 𝑣(∞)
The “natural response” is always an exponential and
satisfies the homogeneous differential equation: 𝑣 = 𝐴𝑒
𝑑𝑣
+𝑣=0
𝑑𝑡
𝑑𝐴𝑒
+ 𝐴𝑒 = 0
𝑑𝑡
𝜆𝐴𝑒 + 𝐴𝑒 = 0
𝜆+1=0
𝜆 = −1
𝑣 = 𝐴𝑒
+𝑣 −
2Ω
0.1 F
8Ω
24 v
6v
𝑖
Complete response = forced + natural
𝑣 = 𝐴𝑒
+ 30
Initial condition: The capacitor is uncharged at
𝑡 = 0 so 𝑣 0 = 0
0 = 𝐴𝑒 + 30
𝐴 = −30
𝑣 = 30(1 − 𝑒 )
First Order Transient
The capacitor is initially uncharged.
1. Find the response for 0 < 𝑡 < 1.
2. Find the response for 𝑡 > 1.
You are expected to know that 3𝑢(𝑡) switches on at 𝑡 = 0, and that 3𝑢 𝑡 − 1 switches on at 𝑡 = 1.
Hence for 0 < 𝑡 < 1 only one source is active, but for 𝑡 > 1, both sources are active.
The capacitor is initially uncharged.
1.Find the response for 0 < 𝑡 < 1.
For 0 < 𝑡 < 1 only one source is active.
The initial condition is 𝑣
= 0.
+𝑣 −
0.1 F
2Ω
8Ω
3A
8Ω
3A
The “final value” of 𝑣 is
𝑣
= −3𝑥8 = 24 volts
The resistance at the capacitor terminals is
𝑅 = 2 + 8 = 10 Ω
The time constant is 𝜏 = 𝑅 𝐶 = 10𝑥0.1 = 1 second.
The “magic formula” for 1st order response is
𝑣 𝑡 =𝑣
+ (𝑣
−𝑣
)𝑒
So
𝑣 𝑡 = 24 + (0 − 24)𝑒
𝑣 𝑡 = 24(1 − 𝑒 ) for 0 < 𝑡 < 1
+𝑣 −
2Ω
−
24
+
2.Find the response for 𝑡 > 1.
For 0 < 𝑡 < 1 the response is
𝑣 𝑡 = 24(1 − 𝑒 )
At 𝑡 = 1, we have
𝑣 1 = 24(1 − 𝑒 ) = 15.17
+𝑣 −
0.1 F
3A
2Ω
8Ω
3A
The “final value” of 𝑣 is
𝑣
= 6 + 24 = 30 volts
+𝑣 −
The “magic formula” does not work because the initial
condition is at 𝑡 = 1 second!
𝑣 𝑡 =𝑣
+ 𝐴𝑒
−
+
where 𝐴 is an arbitrary constant.
24 8 Ω
6
3
A
2
Ω
The “final value” of 𝑣 is 𝑣
= 30 volts
+
−
𝑣 𝑡 = 30 + 𝐴𝑒
At 𝑡 = 1, we have 𝑣 1 = 15.17
15.17 = 30 + 𝐴𝑒
0.3679𝐴 = 15.17 − 30 = −14.83
Check:
𝐴 = −40.31
𝑣 1 = 30 − 40.31𝑒 = 15.17
𝑣 𝑡 = 30 − 40.31𝑒
3A
Second Order Transient
1.Find the initial condition 𝑣 0
2.Find the initial conditions
|
3.Find the differential equation for 𝑣 𝑡 .
4.Find the time constants or the attenuation constant and the
damped natural frequency.
5.Find the final value 𝑣 ∞
6.Find the complete solution for 𝑣 𝑡 .
1.Find the initial condition
For t<0:
When the switch has been in position
1 for a long time:
-the capacitor is an open circuit
-the inductor is a short circuit.
The inductor shorts out the capacitor,
so the initial value is 𝑣 0 = 0
All the current from the source flows
through the inductor.
The initial value of the current in the
inductor is
10
𝑖 0 =
=2
5
2.Find the initial conditions
At t=0, just after the switch throws:
𝑖
The initial value is 𝑣 0 = 0
The initial value of the current in the inductor is
𝑖 0 =2
Since 𝑖 = 𝐶
𝑑𝑣
𝑖
=
𝑑𝑡
𝐶
, we can find
by evaluating
The voltage across 1.1547 ohms is 𝑣 0 = 0 and
so the current in the resistor is 0 amps.
KCL at the top of the circuit gives
−3 − 𝑖 − 𝑖 = 0
𝑖 = −3 − 𝑖 = −3 − 2 = −5 amps
=
=
.
= −20 v/s
3.Find the differential equation for
For t>0:
𝑅 = 1.1547 Ω
𝑣
𝑖
.
KCL for the node at the top of the circuit:
𝑣
−3 − − 𝑖 − 𝑖 = 0
𝑅
𝑑𝑣
𝑖 =𝐶
𝑑𝑡
𝑣
𝑑𝑣
−3 − − 𝐶
−𝑖 =0
𝑅
𝑑𝑡
Differentiate:
1 𝑑𝑣
𝑑 𝑣
𝑑𝑖
+𝐶
+
=0
𝑅 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑣 =𝐿
and 𝑣 =𝑣
1 𝑑𝑣
𝑑 𝑣
𝑣
+𝐶
+ =0
𝑅 𝑑𝑡
𝑑𝑡
𝐿
+
+
𝑣 =0
4.Find the time constants or the attenuation constant
and the damped natural frequency.
For t>0:
𝑅 = 1.1547 Ω
𝑖
.
+
𝑣 =0
The characteristic equation is
𝜆 + 𝜆+ =0
𝑣
The attenuation constant is 𝛼 = 1.732
The damped natural frequency is 𝜔 = 1
The time constant is 𝜏 = =
= 0.5774
+
𝜆 +
𝜆+
=0
.
.
.
𝜆 + 3.464𝜆 + 4 = 0
−3.464 ± 3.464 − 4𝑥4
𝜆=
2
−3.464 ± −4
𝜆=
2
−3.464 ± 𝑗2
𝜆=
2
𝜆 = −1.732 ± 𝑗1
4.Find the time constants or the attenuation constant
and the damped natural frequency.
𝜆 + 3.464𝜆 + 4 = 0
𝜆 = −1.732 ± 𝑗1
𝜆 = −𝛼 ± 𝑗𝜔
The attenuation constant is 𝛼 = 1.732
The damped natural frequency is 𝜔 = 1
The time constant is 𝜏 = =
= 0.5774
.
5.Find the final value
For
The capacitor becomes an open circuit.
The inductor becomes a short circuit.
:
The voltage across the capacitor is zero.
𝑣
𝑣 ∞ =0
𝑖 =0
6.Find the complete solution for
𝑣 𝑡 = 𝐴 𝑒 cos 𝜔 𝑡 + 𝐴 𝑒 sin 𝜔 𝑡
where 𝜏 = 0.5774 and 𝜔 = 1.
𝑣 0 =0
𝑑𝑣
|
= −20 v/s
𝑑𝑡
Set 𝑣 0 = 0:
𝑣 0 = 𝐴 𝑒 cos 0 + 𝐴 𝑒
𝐴 𝑥1 + 𝐴 𝑥0 = 0
𝐴 =0
𝑣 𝑡 =𝐴 𝑒
=− 𝐴 𝑒
sin 0 = 0
sin 𝜔 𝑡
sin 𝜔 𝑡+𝜔 𝐴 𝑒
cos 𝜔 𝑡
| =− 𝐴 sin 0+𝜔 𝐴 cos 0 = −20
𝜔 𝐴 = −20
20
20
𝐴 =−
=−
= −20
𝜔
1
𝑣 𝑡 = −20𝑒
.
sin 𝑡
.
Superposition
Find the voltage across the capacitor.
Remark:
This circuit has a DC source and an AC source,
So we must use Superposition to solve the
circuit.
1.solve the circuit with the DC source
2.solve the circuit with the AC source
3.Add the DC response and the AC response
to get the response to both sources “acting
together”.
8 ohms
10 ohms
8 mF
3 cos(12𝑡)
volts
1H
30 ohms
2H
5 volts
+
𝑣
−
5 mF
1.Solve the circuit with the DC source
8 ohms
At DC or zero frequency, the capacitors
become open circuits and the inductors
become short circuits.
10 ohms
30 ohms
8 ohms
10 ohms
8 mF
3 cos(12𝑡)
volts
+
𝑣
−
1H
5 volts
30 ohms
2H
5 volts
+
𝑣
−
5 mF
The circuit becomes trivially simple!
𝑖=
= 0.1042 amps
𝑣 = 8𝑖 = 8𝑥0.1042 = 0.8333 volts
2.Solve the circuit with the AC source
8 ohms
The DC source is set to zero.
Convert the C’s and L’s to impedance at 𝜔 = 12 rad/sec.
8 mF
1
1
𝑍 = −𝑗
= −𝑗
𝜔𝐶
12𝑥8𝑥10
5 mF
1
1
𝑍 = −𝑗
= −𝑗
𝜔𝐶
12𝑥5𝑥10
1H
𝑍 = 𝑗𝜔𝐿 = 𝑗𝑥12𝑥1 = 𝑗12
2H
𝑍 = 𝑗𝜔𝐿 = 𝑗𝑥12𝑥2 = 𝑗24
10 ohms
8 mF
= −𝑗10.42
= −𝑗16.67
3 cos(12𝑡)
volts
1H
30 ohms
2H
5 volts
+
𝑣
−
5 mF
3 volts
𝑉
8 ohms
Node equations
3−𝑉
𝑉
𝑉 −𝑉
−
−
=0
−𝑗10.42 30 + 𝑗24 10 + 𝑗12
10
𝑉
−𝑗10.42
3−𝑉
𝑉 −𝑉
𝑉
+
−
=0
8
10 + 𝑗12 −𝑗16.67
3
𝑗12
30
𝑗24
+
𝑉
−
−𝑗16.67
Solve the node equations
Eqn 1
3−𝑉
𝑉
𝑉 −𝑉
−
−
=0
−𝑗10.42 30 + 𝑗24 10 + 𝑗12
𝑗0.09597 3 − 𝑉 − (0.02033 − 𝑗0.01626) 𝑉 − 0.04098 − 𝑗0.04918 𝑉 − 𝑉 = 0
−𝑗0.9597𝑉 − 0.02033 − 𝑗0.01626 𝑉 − 0.04098 − 𝑗0.04918 𝑉 + 0.04098 − 𝑗0.04918 𝑉 = −𝑗0.2878
(−0.06131 − 𝑗0.03053) 𝑉 + 0.04098 − 𝑗0.04918 𝑉 = −𝑗0.2878
Eqn 2
3−𝑉
𝑉 −𝑉
𝑉
+
−
=0
8
10 + 𝑗12 −𝑗16.67
0.125(3 − 𝑉) + (0.04098 − 𝑗0.04918)(𝑉 − 𝑉) − 𝑗0.0999𝑉 = 0
−0.125𝑉 + 0.04098 − 𝑗0.04918 𝑉 − 0.04098 − 𝑗0.04918 𝑉 − 𝑗0.05999𝑉 = −0.375
−0.125𝑉 + 0.04098 − 𝑗0.04918 𝑉 − 0.04098 − 𝑗0.04918 𝑉 − 𝑗0.05999𝑉 = −0.375
0.04098 − 𝑗0.04918 𝑉 + −0.16598 − 𝑗0.01081 𝑉 = −0.375
1
𝑉 =
(−0.375 − −0.16598 − 𝑗0.01081 𝑉)
0.04098 − 𝑗0.04918
𝑉 = −3.750 − 𝑗4.500 + 1.530 + 𝑗2.100 𝑉
Eqn 1
(−0.06131 − 𝑗0.03053)( −3.750 − 𝑗4.500 + 1.530 + 𝑗2.100 𝑉) + 0.04098 − 𝑗0.04918 𝑉 = −𝑗0.2878
0.09253 + 𝑗0.3904 + −0.02970 − 𝑗0.1755 𝑉 + 0.04098 − 𝑗0.04918 𝑉 = −𝑗0.2878
0.01128 − 𝑗0.2246 𝑉 = −0.09253 − 𝑗0.6782
𝑉 = 2.991 − 𝑗0.5621 = 3.044∠ − 10.6° hence 𝑣 𝑡 = 3.044 cos(12𝑡 − 10.1°)
Superposition: add the DC response and the AC response.
1.solve the circuit with the DC source
𝑣 = 0.8333 volts
8 ohms
10 ohms
2.solve the circuit with the AC source
𝑣 𝑡 = 3.044 cos(12𝑡 − 10.1°)
By superposition, the complete response is
𝑣 𝑡 = 0.8333 + 3.044 cos(12𝑡 − 10.1°)
8 mF
3 cos(12𝑡)
volts
1H
30 ohms
2H
5 volts
+
𝑣
−
5 mF
Final Exam Tips:
1.Read the whole exam before you start.
2.Do the “easy” questions first.
3.Take your time.
It is better to answer 3 questions correctly than 6 questions
incorrectly.
4.Get the signs right.
5.If the first equation you write is wrong, the
entire solution is wrong.
Good Luck with the Final Exam!