Unformatted text preview: 244 2. 5. Chapter 5 SteadyState 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 rootmeansquare (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
voltagedivision 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 energystorage 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 maximumpower transfer from a twoterminal
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. threephase power
distribution is more economical than single phase.
The power ﬂow in balanced threephase systems
is smooth, whereas power pulsates in singlephase
systems. ...
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 Fall '06
 Preston

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