TEST BANK FOR: For Advanced Practice Psychiatric Nursing, Second Edition: Integrating Psychotherapy, Psychopharmacology, And Complementary And Alternative Approaches Across The Life Span By Kathleen Tusaie Phd Aprn-Bc (Editor) Graded A+.

TEST BANK FOR:
For Electrical Engineering: Principles & Applications 7th Edition
By Allan Hambley (Author)
Questions And Answers Graded A+




1

, Chapter 1

Exercises


E1.1 Charge = Current  Time = (2 A)  (10 S) = 20 C


Dq (T ) D
E1.2 I (T )   (0.01sin(200t)  0.01  200cos(200t )  2cos(200t ) A
Dt Dt

E1.3 Because I2 Has A Positive Value, Positive Charge Moves In The Same
Direction As The Reference. Thus, Positive Charge Moves Downward
In Element C.


Because I3 Has A Negative Value, Positive Charge Moves In The Opposite
Direction To The Reference. Thus Positive Charge Moves Upward In
Element E.


E1.4 Energy = Charge  Voltage = (2 C)  (20 V) = 40 J


Because Vab Is Positive, The Positive Terminal Is A And The Negative
Terminal Is B. Thus The Charge Moves From The Negative Terminal To
The Positive Terminal, And Energy Is Removed From The Circuit
Element.


E1.5 Iab Enters Terminal A. Furthermore, Vab Is Positive At Terminal A.
Thus The Current Enters The Positive Reference, And We Have The
Passive Reference Configuration.


E1.6 Pa (T )  Va (T )Ia (T )  20t 2
(A
)
2

, 10 10
20t 3
10
20t 3
Wa   Pa (T   20t dt 
2   6667 J
3 0
3
)Dt 0

0

(B) Notice That The References Are Opposite To The Passive Sign
Convention. Thus We Have:

Pb (T )  Vb (T )Ib (T )  20t  200
10 10

W  P (T  (20t   10t 2  200t 10  1000 J
)Dt 200)Dt
B  B  0
0 0




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, E1.7 (A) Sum Of Currents Leaving = Sum Of Currents Entering
Ia = 1 + 3 = 4 A

(b) 2 = 1 + 3 + Ib  Ib = -2 A

(c) 0 = 1 + Ic + 4 + 3  Ic = -8 A

E1.8 Elements A And B Are In Series. Also, Elements E, F, And G Are In Series.


E1.9 Go Clockwise Around The Loop Consisting Of Elements A, B, And C:
-3 - 5 +Vc = 0  Vc = 8 V


Then Go Clockwise Around The Loop Composed Of Elements C, D And E:
- Vc - (-10) + Ve = 0  Ve = -2 V


E1.10 Elements E And F Are In Parallel; Elements A And B Are In Series.


E1.11 The Resistance Of A Wire Is Given By R . Using A  D 2 / And
Ρ
 4
A
Substituting Values, We Have:


9.6  1.12  10  L
6
 L = 17.2 M
 (1.6  103 )


E1.12 P V 2 R  R V 2 / P  144   I V /R  120 /144  0.833 A


E1.13 P V 2 R  V  PR  0.25  1000  15.8 V
I V / R  15.8 /1000  15.8 Ma

E1.14 Using Kcl At The Top Node Of The Circuit, We Have I1 = I2. Then, Using
Kvl Going Clockwise, We Have -V1 - V2 = 0; But V1 = 25 V, So We Have V2
= -25 V. Next We Have I1 = I2 = V2/R = -1 A. Finally, We Have
Pr  V2i2  (25)  (1)  25 W And Ps  V1i1  (25)  (1)  25 W.
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