Introduction to Electric Circuits

4 Single-phase a.c. circuits 4.1 ALTERNATING QUANTITIES A quantity which is continually changing its sign from positive to negat ...

4.1 Alternating quantities 67 It follows that f- 1/T (4.1) Example 4.1 Determine (1) the periodic time of an a.c. quantity of fr ...

68 Single-phase a.c. circuits means of a machine called a transformer which, having no moving parts, is extremely efficient. Now ...

Example 4.2 4.1 Alternating quantities 69 Four sinusoidally alternating quantities are represented by: a = 5 sin oot; b = 15 sin ...

70 Single-phase a.c. circuits horizontal position OP the line will have reached position OP 1 after 0~/~o seconds. After 0z/O2 s ...

4.1 Alternating quantities 71 For a sinusoidal current i = I m sin ~ot. Squaring gives i 2 -- Im 2 sin 2 o~t and the 27r, o) mea ...

72 Single-phase a.c. circuits Figure 4.5 N a tf ,.,A A oB b Single-phase a.c. quantities Fig. 4.5 shows an elementary generato ...

4.2 Single-phase a.c. circuits in the steady state 73 by its arrowhead. After the completion of the positive half cycle, of cour ...

74 Single-phase a.c. circuits vt9 1 L Figure 4.8 according to Faraday's law, an emf will be induced in it, given by e = -L(di/d ...

Figure 4.10 XL o q- 4.2 Single-phase a.c. circuits in the steady state 75 I 0 r f and as f "---> ~ SO XL---'~ and I---,0 Exam ...

76 Single-phase a.c. circuits i.e. i = WCVm sin (oJt + 7r/2) Comparing the current and voltage expressions we see that the curre ...

4.3 Series a.c. circuits 77 Solution The capacitive reactance Xc = VII = 100/5 = 20 f~. From Xc = 1/2zrfC we see that the freque ...

78 Single-phase a.c. circuits KVL to the circuit of Fig. 4.14 and taking the clockwise direction as being positive, V- VL- VR = ...

4.3 Series a.c. circuits 79 Solution 1 From Equation (4.15) the inductive reactance X L = toL = 628 • 50 x 10 -3 = 31.4 1) 2 Fro ...

80 Single-phase a.c. circuits As in the case of the RL circuit we can divide each of the phasors in Fig. 4.18(a) by I to obtain ...

Vc = IXc ~l VL = IXL ~~] I (reference) VR = IR (a) VL = IXL 0 Vc = IXc Figure 4.20 VR = IR r (c) 4.3 Series a.c. circuits 81 VL ...

82 Single-phase a.c. circuits Solution 1 The impedance Z = ~//(R 2 + [X g -- Xc]2). Now XL = 2rrfL = 2rr 100 X 150 • 10 -3 -- 94 ...

4.4 Complex notation 83 by the length of the phasor, while Z_~b~ indicates that it is ~b~ degrees ahead of (leading) the referen ...

84 Single-phase a.c. circuits The phasor V3 has the same length as V2 but is 90 ~ ahead of it so that V3 = j V2. But from equati ...

Example 4.10 4.4 Complex notation 85 Represent the following currents on an Argand diagram: (1)/1 = (2 + j3) A (2) I2=(-5+j2) A ...