EE 442: Introduction to Circuits and Instruments
Instructor: Long Que, Ph.D. Email: [email protected]
Office hours (3107 Coover Hall):
Monday 11:00 am - 12:00 pm
Thursday 04:00 pm - 05:00 pm
Friday 09:00 am - 10:00 am
The instructor will usually respond to
Iowa State University
Department of Electrical and Computer Engineering
BB 442 Electric Circuits
Homework 2
Note that no credit will be given to answers not backed by explanations.
1. Find the equivalent resistance between terminals a and b in Figure 1. D
CHAPTER 6
Exercises
E6.1 (a) The frequency of v in (t ) = 2 cos(2 2000t ) is 2000 Hz. For this frequency H (f ) = 260 o. Thus, Vout = H (f )Vin = 260 o 20 o = 460 o
and we have v out (t ) = 4 cos(2 2000t + 60 o ). (b) The frequency of v in (t ) = cos(2 30
CHAPTER 10
Exercises
E10.1 Solving Equation 10.1 for the saturation current and substituting values, we have
Is =
iD exp(vD / nVT ) 1
10 4 = exp(0.600 / 0.026) 1 = 9.502 10 15 A Then for vD = 0.650 V, we have
= 0.6841 mA
iD = I s exp(vD / nVT ) 1 = 9.502
CHAPTER 12
Exercises
E12.1 (a) vGS = 1 V and vDS = 5 V: Because we have vGS < Vto, the FET is in cutoff. (b) vGS = 3 V and vDS = 0.5 V: Because vGS > Vto and vGD = vGS vDS = 2.5 > Vto, the FET is in the triode region. (c) vGS = 3 V and vDS = 6 V: Because
CHAPTER 9
Solutions for Exercises
E9.1 The equivalent circuit for the sensor and the input resistance of the amplifier is shown in Figure 9.2 in the book. Thus the input voltage is
v in = v sensor
Rsensor + Rin
Rin
We want the input voltage with an intern
CHAPTER 11
Exercises
E11.1 (a) A noninverting amplifier has positive gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000t ) (b) An inverting amplifier has negative gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000t ) E11.2
A= v
RL Vo 7
CHAPTER 15
Exercises
E15.1 If one grasps the wire with the right hand and with the thumb pointing north, the fingers point west under the wire and curl around to point east above the wire. If one places the fingers of the right hand on the periphery of th
CHAPTER 13
Exercises
E13.1 given by the Shockley equation: 1 v For operation with iE > I ES , we have exp BE > 1 , and we can write V T v iE I ES exp BE V T Solving for v BE , we have The emitter current is v iE = I ES exp BE V T
i v BE VT ln E I ES
v BC
CHAPTER 14
Exercises
E14.1
(a) iA =
vA RA
iB =
vB RB
iF = iA + iB =
vA RA +
v A vB + RA RB
v o = RF iF = RF
(b) For the vA source, RinA (c) Similarly RinB = RB .
vB RB v = A = RA . iA
(d) In part (a) we found that the output voltage is independent of th
CHAPTER 8
Exercises
E8.1 The number of bits in the memory addresses is the same as the address bus width, which is 20. Thus the number of unique addresses is 220 = 1,048,576 = 1024 1024 = 1024K. (8 bits/byte) (64 Kbytes) = 8 64 1024 = 524,288 bits Startin
CHAPTER 4
Exercises
E4.1 The voltage across the circuit is given by Equation 4.8: v C (t ) = Vi exp( t / RC )
in which Vi is the initial voltage. At the time t1% for which the voltage reaches 1% of the initial value, we have 0.01 = exp( t1% / RC ) Taking
So lmlt'ov
Iowa State University
Department of Electrical and Computer Engineering
BB 442 Electric Circuits
Homework 4
Note that no credit will be given to answers not backed by explanations.
1. For the circuit in Figure I, calculate the value of the unkn
ME 332 Thermodynamics II Homework 8 Due Thursday, April 20th
1. Write the balanced chemical reactions and calculate the equivalence ratios
for methane (CH4), ethane (C2H6), and propane (C3H8) burning with:
a. The theoretical amount of air.
b. Twenty perce
EE 442 Introduction to Circuits and Instruments
Long Que, Ph.D.
Electrical & Computer Engineering
Iowa State University
Email: [email protected]
Introduction
More details can be found in the EE442 Syllabus on Blackboard.
Lectures: Coover 2245; Labs: Coover
EE 442 Midterm
You get one point just for handing in your solution. No points will be given to answers without
explanations. Midterm is close-book exam. Any attempt to communicate during the exam
constitutes academic dishonesty.
Q1. In the circuit in Figu
CHAPTER 1
Exercises
E1.1 E1.2 E1.3 Charge = Current Time = (2 A) (10 s) = 20 C
i (t ) =
dq (t ) d = (0.01sin(200t) = 0.01 200cos(200t ) = 2cos(200t ) A dt dt
Because i2 has a positive value, positive charge moves in the same direction as the reference. Th
CHAPTER 3
Exercises
E3.1
v (t ) = q (t ) / C = 10 6 sin(10 5t ) /(2 10 6 ) = 0.5 sin(10 5t ) V dv i (t ) = C = (2 10 6 )(0.5 10 5 ) cos(10 5t ) = 0.1 cos(10 5t ) A dt
Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t
CHAPTER 5
Exercises
E5.1 (a) We are given v (t ) = 150 cos(200t 30 o ) . The angular frequency is the coefficient of t so we have = 200 radian/s . Then f = / 2 = 100 Hz T = 1 / f = 10 ms Vrms = Vm / 2 = 150 / 2 = 106.1 V Furthermore, v(t) attains a positi
CHAPTER 2
Exercises
E2.1 (a) R2, R3, and R4 are in parallel. Furthermore R1 is in series with the combination of the other resistors. Thus we have: 1 Req = R1 + = 3 1 / R2 + 1 / R3 + 1 / R4
(b) R3 and R4 are in parallel. Furthermore, R2 is in series with
CHAPTER 17
Exercises
E17.1 From Equation 17.5, we have
Bgap = Kia (t ) cos( ) + Kib (t ) cos( 120 ) + Kic (t ) cos( 240 )
Using the expressions given in the Exercise statement for the currents, we have
Bgap = KI m cos(t ) cos( ) + KI m cos(t 240 ) cos( 12
CHAPTER 16
Exercises
E16.1 The input power to the dc motor is Pin = Vsource I source = Pout + Ploss Substituting values and solving for the source current we have 220I source = 50 746 + 3350 I source = 184.8 A Also we have
50 746 Pout 100% = = 91.76% 50 7
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Iowa State University
Department of Electrical and Computer Engineering
BB 442 Electric Circuits
Homework 4
Note that no credit will be given to answers not backed by explanations.
1. For the circuit in Figure I, calculate the value of the u
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Iowa State University
Department of Electrical and Computer Engineering
BB 442 Electric Circuits
Homework 5
Note that no credit will be given to answers not backed by explanations.
1. The circuit in Figure ? has reached its steady state (it
Iowa State University
Department of Electrical and Computer Engineering
BB 201 Electric Circuits
Quiz i}
Note that no credit will be given to answers not backed by explanations. This is an individual assignment,
and any attempt to communicate during the q
i M Git
Quiz 6
Note that no credit will be given to answers not backed by explanations. This is an individual assignment,
and any attempt to communicate during the quiz time constitutes academic dishonesty by deﬁnition (deﬁnition
Iowa State University
Dep
Iowa State University
50 iqu 5, Department of Electrical and Computer Engineering
BB 442 Electric Circuits
Homework 4
Note that no credit will be given to answers not backed by explanations.
l. The circuit in Figure 1 has reached its steady state (it is n