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9 - 244 2 5 Chapter 5 Steady-State Sinusoidal Analysis For...

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Unformatted text preview: 244 2. 5. Chapter 5 Steady-State Sinusoidal Analysis For uniformity, we express sinusoidal voltages in terms of the cosine function. A sine function can be converted to a cosine function by use of the identity sin(z) = cos(: h 90°). The root-mean-square (rms) value (or effective value) of a periodic voltage an) is T v2(r)dr Vrms = — (l The average power delivered to a resistance by nlr) IS vl P IlTlS avg — R Similarly, for a current i (r), we have 1fT‘3mdt T 0 ' and the average power delivered if im flows through a resistance is lrms = Pavg = 13 R rms For a sinusoid. the rms value is the peak value di- vided by $2. We can represent sinusoids with phasors. The mag- nitude of the phasor is the peak value of the sinu- soid. The phase angle of the phasor is the phase angle of the sinusoid (assuming that we have writ— ten the sinusoid in terms of a cosine function). We can add (or subtract) sinusoids by adding (or subtracting) their phasors. The phasor voltage for a passive circuit is the pha- sor current times the complex impedance of the circuit. For a resistance. VR = R13. and the volt- age is in phase with the current. For an inductance, V ,r_ = ijIL, and the voltage leads the current by 90". For a capacitance, V(- = — j ( l/wC)Ic. and the voltage lags the current by 90”. Many techniques learned in Chapter 2 for resis- tive circuits can be applied directly to sinusoidal circuits if the currents and voltages are replaced by phasors and the passive circuit elements are re— placed by their complex impedances. For example, 10. 11. 13. complex impedances can be combined in series or parallel in the same way as resistances (except that complex arithmetic must be used). Node voltages. the current—division principle. and the voltage-division principle also apply to ac circuits. When a sinusoidal current flows through a sinu- soidal voltage. the average power delivered is P = VrmfiirmS cos(9), where 9 is the power angle, which is found by subtracting the phase angle of the current from the phase angle of the voltage (i.e., 6 = 9L. — 6,5). The power factor is cos(9). Reactive power is the flow of energy back and forth between the source and energy-storage elements (L and C). We define reactive power to be pos- itive for an inductance and negative for a capac- itance. The net energy transferred per cycle by reactive power flow is zero. Reactive power is im- portant because a power distribution system must have higher current ratings if reactive power flows than would be required for zero reactive power. Apparent power is the product of rms voltage and rms current. Many useful relationships between power. reactive power. apparent power, and the power angle can be obtained from the power tri- angle shown in Figure 5.22 on page 217. In steady state, a network composed of resistances. inductances. capacitances, and sinusoidal sources (all of the same frequency) has a Thévenin equiva— lent consisting of a phasor voltage source in series with a complex impedance. The Norton equivalent consists of a phasor current source in parallel with the Thévenin impedance. For maximum-power transfer from a two-terminal ac circuit to a load. the load impedance is se- lected to be the complex conjugate of the Thévenin impedance. If the load is constrained to be a pure resistance. the value for maximum power trans- fer is equal to the magnitude of the Thévenin impedance. Because of savings in wiring. three-phase power distribution is more economical than single phase. The power flow in balanced three-phase systems is smooth, whereas power pulsates in single-phase systems. ...
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