Power Computations
2.5 EFFECTIVE VALUES: RMS
34 C H A P T E R 2 Power Computations
(d) When the inductor is energized from the circuit of Fig. 2-5a, the inductor current is described by Eqs. (2-29) and (2-32).
The peak current and peak energy storage are the same as for the circuit of Fig. 2-4a.
The source current has the form shown in Fig. 2-5dand is expressed as
Instantaneous power supplied by the source is
Average source current is zero, and average source power is zero. Peak source power is peak current times voltage, which is 405 W as in part (c).
2.5 Effective Values: RMS 35
Equating the expressions for average power in Eqs. (2-35) and (2-36) gives
or
resulting in the expression for effective or rms voltage
(2-37)
The effective value is the square rootof the meanof the squareof the voltage—
hence the term root mean square.
Similarly, rms current is developed from P I2rmsas
(2-38) The usefulness of the rms value of voltages and currents lies in the computing power absorbed by resistances. Additionally, ac power system voltages and cur- rents are invariably given in rms values. Ratings of devices such as transformers are often specified in terms of rms voltage and current.
IrmsC 1 T3
T
0
i2(t) dt VeffVrmsC
1 T3
T
0
v2(t)dt V2eff
1 T3
T
0
v2(t)dt PV2eff
R 1 RC1
T3
T
0
v2(t)dtS
Figure 2-6 Pulse waveform for Example 2-4.
Vm
DT T t
EXAMPLE 2-4
RMS Value of a Pulse Waveform
Determine the rms value of the periodic pulse waveform that has a duty ratio of Das shown in Fig. 2-6.
36 C H A P T E R 2 Power Computations
■ Solution
The voltage is expressed as
Using Eq. (2-37) to determine the rms value of the waveform gives
yielding
VrmsVm2D VrmsC
1 T3
T 0
v2(t)dtC 1 Ta3
DT 0
V2m dt 3
T DT
02dtbA 1
T (V2mDT) v(t)eVm 0tDT
0 DTtT
EXAMPLE 2-5
RMS Values of Sinusoids
Determine the rms values of (a) a sinusoidal voltage of v(t) Vmsin( t), (b) a full-wave rectified sine wave of v(t) |Vmsin( t)|, and (c) a half-wave rectified sine wave of v(t) Vmsin( t) for 0 tT/2 and zero otherwise.
■ Solution
(a) The rms value of the sinusoidal voltage is computed from Eq. (2-37):
An equivalent expression uses tas the variable of integration. Without showing the details of the integration, the result is
Note that the rms value is independent of the frequency.
(b) Equation (2-37) can be applied to the full-wave rectified sinusoid, but the results of part (a) can also be used to advantage. The rms formula uses the integral of the square of the function. The square of the sine wave is identical to the square of the full-wave rectified sine wave, so the rms values of the two waveforms are identical:
Vrms Vm
12 VrmsF
1
23
2
0
V2m sin2( t) d( t) Vm
12 VrmsF
1 T3
T
0
V2msin2( t)dt where T2
har80679_ch02_021-064.qxd 12/15/09 3:01 PM Page 36
2.5 Effective Values: RMS 37
(c) Equation (2-37) can be applied to the half-wave rectified sinusoid.
The result of part (a) will again be used to evaluate this expression. The square of the function has one-half the area of that of the functions in (a) and (b).
That is,
Taking the 1/2outside of the square root gives
The last term on the right is the rms value of a sine wave which is known to be Vm/ , so the rms value of a half-wave rectified sine wave is
Figure 2-7 shows the waveforms.
VrmsA 1 2
Vm 12Vm
2 12
VrmsaA 1 2bF
1 23
2
0
V2m sin2( t) d( t) VrmsF
1
23
0
V2m sin2( t) d( t) Fa1 2b 1
23
2
0
V2m sin2( t) d( t) VrmsF
1 2£3
0
V2m sin2( t) d( t)3
2
02 d( t)≥F 1 23
0
V2m sin2( t) d( t)
Figure 2-7 Waveforms and their squares for Example 2-5 (a) Sine wave; (b) full-wave rectified sine wave; (c) half-wave rectified sine wave.
i(t) i2(t)
0
(a)
38 C H A P T E R 2 Power Computations
Neutral Conductor Current in a Three-Phase System
An office complex is supplied from a three-phase four-wire voltage source (Fig. 2-8a).
The load is highly nonlinear as a result of the rectifiers in the power supplies of the equip- ment, and the current in each of the three phases is shown in Fig. 2-8b. The neutral cur- rent is the sum of the phase currents. If the rms current in each phase conductor is known to be 20 A, determine the rms current in the neutral conductor.
■ Solution
Equation (2-38) may be applied to this case. Noting by inspection that the area of the square of the current function in the neutral in, is 3 times that of each of the phases ia (Fig. 2-8c)
In, rms F 1 T3
T
0
i2n(t) d(t) F3¢1 T3
T
0
i2a(t) d(t)≤23Ia, rms
Figure 2-7 (continued) i(t)
i2(t)
0
i(t) i2(t)
0
(b)
(c)
EXAMPLE 2-6
har80679_ch02_021-064.qxd 12/15/09 3:01 PM Page 38
2.5 Effective Values: RMS 39
The rms current in the neutral is therefore
Note that the rms neutral current is larger than the phase currents for this situation.
This is much different from that for balanced linear loads where the line currents are In, rms23(20)34.6 A
Figure 2-8 (a) Three-phase source supplying a balanced nonlinear three-phase load for Example 2-8; (b) phase and neutral currents; (c) squares of iaand in.
+ -
+ -
+ - ia
i2a
i2n
van vbn vcn ib
ic
in (a)
(b)
(c) van, ia
vbn, ib
vcn, ic
in
40 C H A P T E R 2 Power Computations
sinusoids which are displaced by 120and sum to zero. Three-phase distribution systems supplying highly nonlinear loads should have a neutral conductor capable of carrying times as much current as the line conductor.
If a periodic voltage is the sum of two periodic voltage waveforms, v(t) v1(t) v2(t), the rms value of v(t) is determined from Eq. (2-37) as
or
The term containing the product v1v2 in the above equation is zero if the functions v1and v2are orthogonal. A condition that satisfies that requirement occurs when v1 and v2 are sinusoids of different frequencies. For orthogonal functions,
Noting that
then
If a voltage is the sum of more than two periodic voltages, all orthogonal, the rms value is
(2-39) Similarly,
(2-40) Note that Eq. (2-40) can be applied to Example 2-6 to obtain the rms value of the neutral current.
Irms2I21,rmsI22,rmsI23,rms Á Ba
N n1
I2n,rms Vrms2V21,rmsV22,rmsV23,rms Á Ba
N n1
V2n,rms
Vrms2V21,rmsV22,rms
1 T3
T
0
v21(t) dt V21,rms and 1 T3
T
0
v22(t) dt V22,rms V2rms1
T3
T
0
v21(t) dt1 T3
T
0
v22(t)dt V2rms1
T3
T
0
v21 dt1 T3
T
0
2v1v2 dt1 T3
T
0
v22 dt V2rms1
T3
T
0
A
v1v2B
2 dt1T3T
0
A
v212v1v2v22B
dt13 har80679_ch02_021-064.qxd 12/15/09 3:01 PM Page 40
2.5 Effective Values: RMS 41
RMS Value of the Sum of Waveforms
Determine the effective (rms) value of v(t) 4 8 sin( 1t10) 5 sin ( 2t50) for (a) 22 1and (b) 2 1.
■ Solution
(a) The rms value of a single sinusoid is Vm/ , and the rms value of a constant is the constant. When the sinusoids are of different frequencies, the terms are orthogonal and Eq. (2-39) applies.
(b) For sinusoids of the same frequency, Eq. (2-39) does not apply because the integral of the cross product over one period is not zero. First combine the sinusoids using phasor addition:
8⬔10 5⬔50 12.3⬔25.2 The voltage function is then expressed as
v(t) 4 12.3 sin ( 1t25.2) V The rms value of this voltage is determined from Eq. (2-39) as
VrmsC42a 12.3
12b29.57 V Vrms2V21,rmsV22,rmsV23,rmsC42a
8
12b2a 5
12b27.78 V 12
EXAMPLE 2-7
EXAMPLE 2-8
RMS Value of Triangular Waveforms
(a) A triangular current waveform like that shown in Fig. 2-9a is commonly encountered in dc power supply circuits. Determine the rms value of this current.
(b) Determine the rms value of the offset triangular waveform in Fig. 2-9b.
■ Solution
(a) The current is expressed as
The rms value is determined from Eq. (2-38).
I2rms1 TC
3
t1
0
¢2Itm
1
tIm ≤2dt 3
T
t1
¢T2Itm
1
tIm(Tt1) Tt1 ≤2dtS i(t)μ
2Im
t1 tIm 0tt1 2Im
Tt1tIm(Tt1)
Tt1 t1tT
42 C H A P T E R 2 Power Computations
The details of the integration are quite long, but the result is simple: The rms value of a triangular current waveform is
(b) The rms value of the offset triangular waveform can be determined by using the result of part (a). Since the triangular waveform of part (a) contains no dc component, the dc signal and the triangular waveform are orthogonal, and Eq.
(2-40) applies.