0195136047.pdf

54 CIRCUIT CONCEPTS 1.1.5A wire withn= 1030 electrons/m^3 has an area of cross sectionA=1mm^2 and carries a current i=50 mA. Com ...

PROBLEMS 55 ord^2 cmil. The handbook for aluminum electrical conductors lists a dc resistance of 0.01558per 1000 ft at 20°C for ...

56 CIRCUIT CONCEPTS the internal resistance of the amplifier isRS= 8 , one has connected a mismatched speaker with RL= 16 . De ...

PROBLEMS 57 would not drop by more than 2% with respect to the source open-circuit voltage. *1.2.12A practical current source is ...

58 CIRCUIT CONCEPTS (a) 3 Ω 6 Ω Ix Ix 1 Ω Ix 1 Ω 1 Ω 2 Ω 2 Ω 6 A 10 A (c) 3 Ω 2 Ω^2 Ω 4 Ω Ix (b) (d) 2 Ω 3 Ω 5 Ω 6 A 4 Ω 2 Ω 8 A ...

PROBLEMS 59 1.2.24Determine the equivalent capacitance at terminals A–Bfor the circuit configurations shown in Figure P1.2.24. 1 ...

60 CIRCUIT CONCEPTS 10 20 30 − 1 1 (20t + 10) (− 10 t + 40) 10 t 30 10(t + 1) 0 L = 100 pH 23 + − i(t), A i(t) v(t) t, μs 45 Fig ...

PROBLEMS 61 L 11 L 22 i 1 i 2 v 1 v 2 + − + − M (a) L 11 L 22 i 1 i 2 v 1 v 2 − + − + M (b) Figure P1.2.32 i 2 i 1 v 2 v 1 v 3 L ...

62 CIRCUIT CONCEPTS termine the remaining voltages and currents. Also calculate the power delivered to each element as well as t ...

PROBLEMS 63 3 V iy vCF iz ix + − + − + −+ +−vAB −1 V −2 V +− − 1 V + − F A B C D H v =? i =? G 6 A E − + 4 V 3 A 2 A S Figure P1 ...

64 CIRCUIT CONCEPTS + − 5 A 5 Ω 10 V Figure P1.3.7 + − + − IS 5 Ω 10 Ω 5 10 Ω V^0 = 10 V V 1 Figure P1.3.8 + − + − 1 F 1 F 1 H S ...

PROBLEMS 65 R 2 R 4 Vin V 1 R 1 + +− − R 3 Figure P1.4.5 R 4 =? RM = 6 Ω b a R 1 = 16 Ω R 2 = 8 Ω 12 Ω + − R 3 = 40 Ω 24 V G Fig ...

2 Circuit Analysis Techniques 2.1 Thévenin and Norton Equivalent Circuits 2.2 Node-Voltage and Mesh-Current Analyses 2.3 Superpo ...

2.1 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 67 whereAandBare two constants. The Thévenin equivalent circuit at any two terminals ...

68 CIRCUIT ANALYSIS TECHNIQUES a b 24 Ω 48 Ω 6 A (a) 96 V I =? + − R = 16 Ω a Voc b IL 48 Ω 24 Ω (b) 96 V 144 V + − + + − − a b ...

2.1 THÉVENIN AND NORTON EQUIVALENT CIRCUITS 69 (a) KVL: 144− 24 IL− 48 IL− 96 =0, or 72IL=48, orIL=^2 / 3 A Voc= 144 − 24 (^2 / ...

70 CIRCUIT ANALYSIS TECHNIQUES a b I 200 Ω 9 I 2000 Ω + − (a) I 1 10 V Figure E2.1.2 a b Isc I 200 Ω 9 I 2000 Ω + − (b) I 1 10 V ...

2.2 NODE-VOLTAGE AND MESH-CURRENT ANALYSES 71 connected to a receptacle (plug or outlet) in a much simpler model with the open-c ...

72 CIRCUIT ANALYSIS TECHNIQUES R 3 R 1 R 2 Node O Node A Node B V 1 + − V 2 + − Figure 2.2.1Circuit for illustration of nodal-vo ...

2.2 NODE-VOLTAGE AND MESH-CURRENT ANALYSES 73 EXAMPLE 2.2.1 By means of nodal analysis, find the current delivered by the 10-V s ...