**Solutions****Manual**of**Electric****Circuits****SOLUTIONS****MANUAL**Fundamentals of**Electric***SOLUTIONS**MANUAL**Electric**Circuits*9th Ed by*Solutions**Manual*for*Electric*Circuit Analysis

**Solutions** **Manual** of **Electric** **Circuits**

Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..." Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources, then" between the "(c)" and "Connect a 1-A source. ." The edited phrase will read: "Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Then Rt = Vab/1." Page 340, Problem P8.3-5: The answer should be Page 340, Problem P8.3-6: The answer should be . Page 341, Problem P.8.4-1: The answer should be Page 546, line 4: The angle is instead of .

**SOLUTIONS** **MANUAL** Fundamentals of **Electric**

Chapter 6: Inductance, Capacitance, and Mutual Inductance 7. Chapter 10: Sinusoidal Steady-State Power Calculations You may use the homework *solutions* provided below.

*SOLUTIONS* *MANUAL* *Electric* *Circuits* 9th Ed by

1.3-4 dq ( t ) i (t ) = dt 0 i (t ) = 2 -2( t - 2 ) -2e t 57.7 s 0 3-11 P3.4-6 240 a.) 18 = 12 V 120 240 18 b.) 18 = 0.9 W 120 240 R c.) 18 = 2 18 R = 2 R 2 (120 ) R = 15 R 120 R d.) 0.2 = ( 0.2 )(120 ) = 0.8 R R = 30 R 120 (checked using LNAP 8/16/02) 3-12 Section 3-5 Parallel Resistors and Current Division P3.5-1 i = 1 i = 2 i = 3 i = 4 1 1 1 6 4= 4= A 1 1 1 1 1 2 3 6 3 6 3 2 1 1 2 3 4 = A; 1 1 1 1 3 6 3 2 1 1 2 4 =1 A 1 1 1 1 6 3 2 1 1 4=2 A 1 1 1 1 6 3 2 P3.5-2 (a) (b) (c) 1 1 1 1 1 = = R = 2 R 6 12 4 2 v = 6 2 = 12 V p = 6 12 = 72 W P3.5-3 i= 8 8 or R1 = R1 i 8 8 or R 2 = 2-i R2 8 = R 2 (2 - i ) i = 2 - (a) (b) 8 4 8 = A ; R1 = =6 4 12 3 3 8 2 8 i = = A ; R2 = =6 2 12 3 2- 3 i = 2- 3-13 ( c ) R1 = R 2 2 P3.5-4 R1 R 2 R1 R 2 1 will cause i= 2 = 1 A. 2 1 = 8 ; R1 = R 2 2 R1 = 8 R1 = 8 R1 = R 2 = 8 2 Current division: 8 i = -6 = -2 A 1 16 8 ( ) 8 i = -6 = -3 A 2 8 8( ) i = i -i 1 2 P3.5-5 = 1 A R 1 i and current division: i = 2 R R s 1 2 Ohm's Law: v = i R yields o 2 2 v R R 2 i = o 1 s R R 2 1 plugging in R = 4, v R1 then vo R2 R4 = 1 vs R1 R2 R3 R2 R vo R - 2 = 1 and - 1 4 R1 R2 R2 vs R3 6-1 Ex. 6.61 R vin - vout vin 0 = 0 vout = 1 f vin Rf R1 R1 when R f = 100 k and R1 = 25 k then 100 103 vout = 1 = 5 vin 25 103 6-2 Ex.

*Solutions* *Manual* for *Electric* Circuit Analysis

We will follow the book, **Electric** **Circuits** (10th edition) by Nilsson and Riedel.

Introduction to electric circuits solution manual:

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