Circuit Variables
1
Assessment Problems
AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientic notation: $100 billion = $100 109 Now we determine the
Circuit Elements
2
Assessment Problems
AP 2.1
[a] To nd vg write a KVL equation clockwise around the left loop, starting below the dependent source: ib ib so vg = + vg = 0 4 4 To nd ib write a KCL equation at the upper right node. Sum the currents leaving
Simple Resistive Circuits
3
Assessment Problems
AP 3.1
Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 resistor and the 10 resistor in
4 Techniques of Circuit Analysis
Assessment Problems
AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:
The two node voltage equations are v1 v1 v2 v1 + + =0 15 + 60 15 5 v2 v2 v1 5+ + =0 2 5 Place these equations in sta
The Operational Amplier
5
Assessment Problems
AP 5.1 [a] This is an inverting amplier, so vo = (Rf /Ri )vs = (80/16)vs , vs ( V) 0.4 2.0 so vo = 5vs
3.5 0.6 1.6 2.4
vo ( V) 2.0 10.0 15.0 3.0 8.0 10.0 Two of the vs values, 3.5 V and 2.4 V, cause the op amp
6 Inductance, Capacitance, and Mutual
Inductance
Assessment Problems
AP 6.1 [a] ig = 8e300t 8e1200t A v=L dig = 9.6e300t + 38.4e1200t V, dt 38.4e1200t = 9.6e300t t > 0+
v (0+ ) = 9.6 + 38.4 = 28.8 V [b] v = 0 when or t = (ln 4)/900 = 1.54 ms [c] p = vi =
7 Response of First-Order RL and RC
Circuits
Assessment Problems
AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 resistor from the circuit.
First combine the 30 and 6 resi
10 Sinusoidal Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/ 45 V, I = 20/15 A Therefore 1 P = (100)(20) cos[45 (15)] = 500 W, 2 Q = 1000 sin 60 = 866.03 VAR, [b] V = 100/ 45 , I = 20/165 BA AB
AB
BA
P = 1000 cos(210 ) = 866.03 W
11 Balanced Three-Phase Circuits
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to nd VBC . To do this, write a KVL equation to nd VAB , and use the known phase angle relationship between VAB and VBC to nd VBC . VAB = VAN + VNB = VAN VBN
12 Introduction to the Laplace Transform
Assessment Problems
AP 12.1 [a] cosh t = et + et 2 Therefore, 1 (s )t [e + e(s+ )t ]dt Lcfw_cosh t = 2 0 = = [b] sinh t = 1 e(s )t 2 (s ) 1 2
0
+
e(s+ )t (s + ) = s2
0
1 1 + s s+
s 2
et et 2 Therefore, 1 (s )t e
14
Introduction to Frequency-Selective Circuits
Assessment Problems
AP 14.1 fc = 8 kHz, c = 1 ; RC c = 2fc = 16 krad/s R = 10 k;
. C =
1 1 = = 1.99 nF c R (16 103 )(104 )
AP 14.2 [a] c = 2fc = 2 (2000) = 4 krad/s L= 5000 R = = 0.40 H c 4000 4000 c = c + j
Active Filter Circuits
15
Assessment Problems
AP 15.1 H (s) = (R2 /R1 )s s + (1/R1 C ) R1 = 1 , . C = 1 F
1 = 1 rad/s; R1 C R2 = 1, R1 .
. R2 = R1 = 1 s s+1
Hprototype (s) =
AP 15.2 H (s) =
20,000 (1/R1 C ) = s + (1/R2 C ) s + 5000 C = 5 F
1 = 20,000; R1
Fourier Series
16
Assessment Problems
AP 16.1 av = 1 T 2 T
2T /3 0
Vm dt +
1 T
T 2T /3
Vm 3
T
7 dt = Vm = 7 V 9 Vm cos k0 t dt 3
ak = = bk = =
2T /3 0
Vm cos k0 t dt + sin 4k 3 =
2T /3
4Vm 3k0 T 2 T
2T /3 0
6 4k sin k 3
T 2T /3
Vm sin k0 t dt + 1 cos 4k 3
A (B C) = B(A C) C(A B) (A B) C = B(A C) A(B C) ( A B) ( A + B) = A 2 B2 (A B) (A + B) = 2A B (A B) (C D) = (A C)(B D) (A D)(B C) (A B) (C D) = (A B D)C (A B C)D (A B) (A B) = (A B)2 = A2 B2 (A B)2 (A B) (B C) (C D) = (A B C)2 A ( B C ) + B ( C A ) + C (
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CHAPTER 6
Laplace Transform
Major Changes
This chapter underwent some major changes regarding the order of the material and the emphasis placed on the various topics. In particular, Diracs delta was placed in a separate