PROBLEM # 6.27
Find the equivalent capacitance with respect to the terminals ab for the circuit
shown.
1
1 1 5
= + =
Ceq1 4 6 12
Ceq1 =2.4 m F .12V
Ceq2 =1.6 m F +2.4 m F =4 m F .12V
1
1 1 16 1
= + = =
Ceq3 4 12 48 3
Ceq3 =3m F .7V
Ceq4 =5m F +3m F =8m F
PROBLEM # 6.3
The voltage at the terminals of the 200H inductor is shown below in (b). The inductor
current is known to be zero for t<=0.
a) Determine the expressions for i for t>=0.
vs (t ) =0, t 0
vs (t ) =5mV , 0 t 2ms
vs (t ) =0, t 2ms
t
1
i (t ) = v(
PROBLEM # 6.28
Use realistic capacitor values from Appendix H to construct series and parallel
combinations of capacitors to yield the equivalent capacitances specified below.
Try to minimize the number of capacitors used. Assume that no initial energy is
Problem # 4.91
a) Use the principle of superposition to find the voltage v in the circuit shown.
b) Find the power dissipated in the 10 resistor.
Consider the voltage source acting alone.
Req (10 (2 12)
(10)(14)
5.833
10 14
v
v'
5.833
5.833
(110V ) 59.2
Problem # 4.93
Use the principle of superposition to find the voltage v 0 in the circuit shown.
The 240 V source acting alone.
'
0
v
(20 5)
(20 5) 12
(240V ) 60V
The 84 V source acting alone
"
0
v
(20 12)
(20 12) 5
( 84V ) 50.4V
The 16 A source acting a
ECE 201
Exam #2
Review
Topics
Thevenins/Nortons Theorem
Maximum Power Transfer
Superposition
OP AMPs
Inverting Amplifier
Non Inverting Amplifier
Relaxation Oscillator
Comparator
Difference Amplifier
Inductance
Henry (H)
mH, H
Magnetic Field
Lines of Flux
Problem # 4.6
Use the nodevoltage method to determine the voltages v 1 and v2 in the circuit
shown below.
The voltages have been identified for you show the ground connection
(voltage=0) and the currents out of the nodes.
The direction of the 3A current
PROBLEM # 6.32
Determine the equivalent circuit for a series connetion of ideal capacitors. Assume that
each capacitor has its own initial voltage.
The current in each capacitor is the same since they are in series and their individual
voltages add togeth
PROBLEM # 6.42
The polarity markings on two coils are to be determined experimentally. The
experimental setup is shown below. Assume that the terminal connected to the negative
terminal of the battery has been given a ploarity mark as shown. When the swit
PROBLEM # 6.40
a) Show that the coupled coils shown below can be replaced by a single coil having an
inductance of Lab = L1 + L2 + 2M.
b) Show that if the connections to the terminals of the coil labeled L2 are reversed,
Lab = L1 + L2 2M.
vab
a)
+

iab
v
Superposition
Principle of Superposition
When a linear circuit is excited by more than one
independent source of energy, the total response is
the sum of the responses to each source acting
individually.
To find the response due to each independent
sou
Thevenin and Norton Equivalent
Circuits
Voltage Source Model
Current Source Model
ECE 201 Circuit Theory I
1
Why do we need them?
Circuit simplification
Reduce the complicated circuit on the left to a
voltage source in series with a resistor.
ECE 201 Cir
Maximum Power Transfer
Maximize the power delivered to a
resistive load
ECE 201 Circuit Theory I
1
Consider the General Case
A resistive network contains independent
and dependent sources.
A load is connected to a pair of terminals
labeled a b.
What va
PROBLEM #7.55
Assume that the switch in the circuit shown has been in position a for a long time
and that at t = 0 it is moved to position b.
Find
a) vC(0+)
vC (0+ =vC (0 ) =50V
)
b) vC()
20
( 30)
20 +5
vC ( ) = 24V
vC ( ) =
c) for t > 0
t =Req C =( 2
PROBLEM # 7.35
The switch in the circuit shown has been closed for a long time before opening at
t = 0.
a) find the numerical expressions for iL(t) and vO(t) for t>=0.
For t<0, the switch is closed, and the 4 mH inductor appears as a short circuit.
The 5
PROBLEM # 7.81
The voltage waveform is applied to the circuit shown. The initial voltage on the
capacitor is zero.
a) Calculate vo(t).
vs (t ) =0, 0 t
vs (t ) =50V , 0 t 1ms
vs (t ) =0,1ms t
For 0 < t < 1ms
vC (0) =0V
vC ( ) =50V
for t > 1ms
t =RC =(400 1
PROBLEM # 7.34
a) Use component values fromAppendix H to create a firstorder RL circuit (see
Fig. 7.16) with a time constant of 8 s. Use a single inductor and a network of
resistors, if necessary. Draw your circuit.
L
R
L
R=
t
L =1mH
t=
1 10 3
R=
=125W
Alternator Problem
1.)
Identify the branch currents and node A.
2.)
Write equations for the branch currents and solve.
A
VA
+
I3
I2
I1
+

16 VA
I1
0.2
12.8 VA
I2
0.1
16 VA 12.8 VA
VA
0.2
0.1
16 VA 25.6 2VA 0.2VA
41.6 3.2VA
VA 13Volts
I3
VA
1
3.)
Calcu
UNIVERSITY OF MASSACHUSETTS DARTMOUTH
DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING
ECE 201
CIRCUIT THEORY I
ALTERNATOR PROBLEM
An automobile alternator with an internal resistance of 0.2 develops an open
circuit voltage of 16.0 Volts. The storage bat
7.2: Definition of the Laplace Transform
2
2. L cfw_t (s) = 2/s3.
3t
4. L cfw_te (s) = 1/(s3)2.
s
10. L cfw_f (t) (s) = e /s2 + 1/s 1/s2.
12. L cfw_f (t) (s) = (1e
3(s2)
)/(s2) + e
3s
6
/s, for s2. For s=2, L cfw_f (t) (2) = 3 + e /2
2t 2
3
14.
7.3: Properties of the Laplace Transform
3
4. L cfw_3t42t2+1=72/s5 4/s +1/s, is valid for s > 0.
6. L cfw_e2t sin 2t + e3tt2=2/(s+2)2+4) + 2/(s3)3, s >3.
8. L cfw_(1+et)2=1/s + 2/(s+1) +1/(s+2), s>0.
12. Lcfw_sin3t cos3t = Lcfw_sin6t/2 =3/(s2+36).
2
30.
4.4: Nonhomogeneous Equations: The Method of Undetermined
Coefficients
t
(ln 3)t
rt
4. Rewriting the righthand side in the form 3 = e
= e , where r = ln 3, we
conclude that the method of undetermined coefficients can be applied.
10. y (t) 3 is a particul
2.2: Separable Equations
2. This equation is separable because we can separate variables by multiplying both sides
by dx and dividing by 4y2 3y+1.
4. This equation is separable because
dy/dx= yex+y / (x2 +2)= [ex /(x2 +2)] yey =g(x)p(y).
6. Writing the eq
3.2: Compartmental Analysis
2. Let x(t) denote the mass of salt in the tank at time t with t = 0 denoting the moment
when the process started. Thus we have x(0) = 0.5 kg. We use the mathematical model
described by equation (1) of the text to find x(t). Si
7.6: Transforms of Discontinuous and Periodic Functions
6. The function g(t) equals zero until t reaches 2, at which point g(t) jumps to t + 1. We write
g(t) = 00,2(t)+(t+1)u(t2) = (t+1)u(t2)
so then
Lcfw_g(t) = e
8. Since g(t)=(sint)u(t/2), then Lcfw_g(t