Instantaneous Power in AC Circuits

Instantaneous Power in AC Circuit

Theoretical Background Starting from the main principles, it is well known that the potential energy (E) is "the summation of the work ? " done to achieve the energy as described mathematically by the formula: ? = ? .(1) The power ? defined as "the rate of change of work" or "the change of work with respect to time", hence it can be described as: ? ? ..(2) ? = Since the small change in any quantity denoted by the symbol and the infinitismal change by the symbol ? , then, the instantaneous value of the variable power ? can be described as: ? = ? or ? =?? ?? ...(3) ?,

As, the constant electric current ? flowing in a conductor is defined as; "the change in electric charge ? with respect to time", hence it can be described as: ? =? ? .. ..(4) Similarly, the instantaneous value of the variable current is ? and it can be described as: ? = ? ?=?? ?? .(5) As, ? is the instantaneous value of ? as well. The constant electrical voltage ? generated by certain source is defined as "the work done per unit charge", hence it can be described as: ? =? ? (6)

Similarly, the instantaneous value of the variable voltage is ? and it can be described as: ? = ? ? or ? =?? ?? (7) Then, the constant electrical power ? can be considered as ? = ? ? Since, ? = ? ? because: and ? = ? ? , then it can be concluded that (? = ??), ? ? ? ? ?= ? .(8) ?= In the same manner the instantaneous power will be: ? = ?? .(9) Since, ?? ?? ?? ?? ??= ? ..(10) ?? = ??=

Instantaneous and Average AC Power: Let the instantaneous value of the supplied voltage from an AC source to certain circuit being ? ? = ??sin (?? + ??) And the instantaneous current supplied from that AC source to this circuit being ? ? = ??sin (?? + ??) Then the instantaneous power supplied by that source being ? ? = ? ? ?(?) ? ? = ??sin (?? + ?? ??sin (?? + ?? Now, since sin ? sin (?) =cos ? ? cos (? + ?) 2

Then ? ? =???? cos ?? + ?? ?? ?? cos (?? + ??+ ?? + ??) 2 ? ? =???? cos ?? ?? cos (2?? + ??+ ??) 2 ? ? = ????????cos ?? ?? cos (2?? + ??+ ??) ? ? = ????????cos ?? ?? ????????cos2?? + ??+ ?? .(11) Then it is quite clear that the instantaneous power formula composed of two terms, the first is constant with time while the second varies sinusoidally at double frequency with time and its average value through one cycle is zero. To understand the equation (11), let us consider the voltage waveform as a reference, and this leads to the fact that (??= 0). Also, let us consider the following cases:

Pure Resistive Load: In this case, the source voltage and current are in phase, then (??= 0) w.r.t. voltage. Hence, equation (11) becomes: ? ? = ????????cos0 ????????cos (2??) ? ? = ???????? ????????cos (2 t) (12) Then it is clear that instantaneous value composed of two components. One of them is constantly equal to VrmsIrms, while the second term is of cosinusoidal waveform superimposed on the constant term. The average value of the second term is equal to zero. Then "the average power transferred from the source to the load in this case equal to???????? only". Figure (1) represents the instantaneous real power waveform and its average and maximum values.

Fig. (1) The resistive (real power) waveform

Figure (1) shows that the instantaneous value of the real power has always been positive. This is due to the fact that the source current and the source voltage are in phase always. This means that the source voltage and current are at the same polarity always in this case. It leads to the fact that the source produces the power and transfer it to the load. The resistive load, voltage and current waveforms are out of phase by 180 always. This means that the polarity of the voltage across the resistance and that of the current passing through it are reversed always, so the resistance absorbed power all the time. So, it is called an (Active Power). Pure Inductive Load: In this case, the source current lags the source voltage by ninety degrees, then (??= 90?) . Hence, equation (11) becomes: ? ? = ????????cos (+90? ) ????????cos (2?? 90?)

Since, in general cos ? 90?= sin (?), then ? ? = ????????sin (2??) ? ? ????????? = ????????sin2?? .(13) Then it is clear that "the average value of the power transferred from the source to the load equal to zero", and it varies instantaneously with time as a negative sinusoidal waveform of double frequency. Figure (2) represents the instantaneous inductive (reactive) power waveform. It is clear that the positive part of the power occurs when the source voltage and source current are in the same polarity, while it will be negative when they are reversed. This leads to the truth states that the source in an AC circuit will act as a load for during the half time of one cycle, hence it means that the load will act as a source for the other half time of one cycle.

It may be considered as a strange truth, but it can be easily explained physically, if the Lenz's law is considered. This law deals with the direction of any induced current. In this case, a current induced in the coil winding affected by the change in the flux generated due to the source current passing through the coil turns. This law states that "The direction of the induced current will be such as to oppose the cause producing it". Hence, the induced current in the coil changes its direction (polarity) due to the change in the flux rate of change (the cause of induction) in each quarter cycle. This induced current leads to generate a voltage across the coil equal to that of the source and out of phase with it be 180 . Then, in such a circuit there are two sources and two loads acts alternatively at each quarter cycle, leading the power to go (forward) and return (backward) through the circuit.

Fig. (2) The inductive (reactive power) waveform

Pure Capacitive Load: In this case, the source current leads the source voltage by ninety degrees, then (??= +90?) . Hence, equation (11) becomes: ? ? = ????????cos ( 90? ) ????????cos (2?? + 90?) Since cos ? + 90?= sin (?), then ? ? = +????????sin (2??) ? ? ?????????? = +????????sin2?? ..(14) Then it is clear that "the average value of the power transferred from the source to the load equal to zero", and it varies instantaneously with time as a positive sinusoidal waveform of double frequency. Figure (3) represents the instantaneous capacitive (reactive) power waveform. The same behavior to that of the inductive load will occur, but in out of phase. This is easy to be explained due to the charging and discharging process of the capacitive load. It is clear that the operation of the capacitor is of opposite timing.

Fig. (3) The capacitive (reactive power) waveform.

From the previous analysis it is clear that there are two types of power in AC circuits: 1. A power transfers from the source to the load during all the time of the cycle. This power due to the existence of the resistive part and is called the real power. Hence, it is an oscillatory unidirectional part of the AC power. 2. The power that transfer alternatively from source to load and from load to source at each quarter cycle, hence it oscillate at double frequency of the applied voltage. This is called the imaginary power. Hence, it is a bidirectional part of the AC power. The imaginary power occur in the AC circuit due to: A. The inductive load, so it is an inductive power. This power oscillates in a sinusoidal waveform in double the frequency of the applied voltage. B. The capacitive load, so it is a capacitive power. This power oscillates in a sinusoidal waveform of double the frequency of the applied voltage, but it is in out of phase with that of the inductive load. C. If both are exist then, the resultant imaginary or reactive power can be determined by adding these two components due to the formula: ? ? ????????= ? ? ??????????+ ? ?????????? In general: The total instantaneous AC power in an (? ? ?) circuit (the apparent power) composed of an oscillatory unidirectional part superimposed by an oscillatory part of a resultant imaginary power as shown in the equation below. ? ?????????= ? ???????+ ? ????????? Or ? ?????????= ? ?????+ ? ??????????

Ex.1: For the circuit shown below determine the instantaneous apparent (total) power, assume that the sinusoidal waveform current of frequency= (50Hz) passing through the circuit and of an rms value= 2A, by summation of the instantaneous power on each element. The total instantaneous power fed to the circuit can be determined as follows: ? = 10 ??= 10 ??= 30 ? ??= ??? + ??? + ??? ?? By applying equations (12, 13, 14) to determine [??? , ??? , ??? ] consequently. The following waveforms show each one of these powers as a function of time

Inst. Power in Resistance 45 40 35 30 Power (W) 25 20 P res =P real 15 10 5 0 0 2 4 6 8 10 12 14 16 18 20 Time (ms) ??? = ????????? ?????????cos 2?? ?????= ? ????

Inst. Power in Res. and Inductance 50 40 30 20 Power (W) 10 P res =P real Pind 0 0 5 10 15 20 -10 -20 -30 Time (s) ??? = ?????????sin 2?? ?????= ??????

Inst. Power in R, L, and C 80 60 40 Power (W) 20 P res =P real 0 0 2 4 6 8 10 12 14 16 18 20 Pind Pcap -20 -40 -60 -80 Time (s) ??? = ?????????sin 2?? ?????= ??????

Determination of Inst. Reactive Power 80 60 40 Power (W) 20 P res =P real 0 Pind 0 2 4 6 8 10 12 14 16 18 20 Pcap -20 Preac=Pimag -40 -60 -80 Time (s) ?????????? = ??? + ???

All Power Components 80 60 40 Power (W) 20 P res =P real Pind 0 Pcap 0 2 4 6 8 10 12 14 16 18 20 Preac=Pimag -20 P appparent -40 -60 -80 Time (s) ? ??= ??? + ??? + ???

The Main Three Types of Power 50 40 Instantaneous Power (W) 30 20 P real 10 P reactive P apparent 0 0 2 4 6 8 10 12 14 16 18 20 -10 -20 -30 Time (ms)

The Power Triangle Axis of the average real power- ??? ???? CCW 50 40 30 20 P real 10 P reactive P apparent 0 0 5 10 15 20 -10 -20 -30 ? = ? + ?? ? = ???? ???? CCW ? = ? cos? = ???? ???? cos? ? = ? sin? =???? ???? sin? ? = tan 1? Imaginary or Reactive Power ? (VAr) Apparent Power ? (VA) ? ? Real or Active Power ? (W) This is known as a power triangle: It is a phasor diagram representing the instantaneous value of the complex power. It rotates CCW around a center of (0.0, +??? ???? .

t 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 0.005 0.0055 0.006 0.0065 0.007 0.0075 0.008 0.0085 0.009 0.0095 0.01 0.0105 0.011 0.0115 0.012 0.0125 0.013 0.0135 0.014 0.0145 0.015 0.0155 0.016 0.0165 0.017 0.0175 0.018 0.0185 0.019 0.0195 0.02 t (ms) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18 18.5 19 19.5 20 P res =P real 0 0.978868 3.819654 8.244282 13.81964 19.99997 26.18031 31.75567 36.18031 39.02112 40 39.02115 36.18038 31.75576 26.18041 20.00008 13.81974 8.244368 3.819716 0.978901 2.82E-10 0.978835 3.819591 8.244196 13.81954 19.99987 26.18021 31.75559 36.18025 39.02108 40 39.02118 36.18044 31.75585 26.18051 20.00019 13.81984 8.244454 3.819779 0.978934 1.13E-09 Pind 0 6.18033 11.7557 16.1803 19.0211 - 20 19.0211 16.1804 11.7557 6.18039 -5.3E-05 6.180284 11.75565 16.1803 19.02111 20 19.02116 16.18039 11.75578 6.180436 0.000106 6.18023 11.7556 16.1803 19.0211 - 20 19.0212 16.1804 11.7558 6.18049 0.00016 6.180183 11.75557 16.18024 19.02107 20 19.02119 16.18046 11.75587 6.180537 0.000212 Pcap 0 18.541 35.26709 48.54099 57.06337 60 57.06342 48.54109 35.26722 18.54116 0.000159 18.5409 35.267 48.5409 57.0633 - 60 57.0635 48.5412 35.2673 18.5413 0.00032 18.5407 35.26683 48.5408 57.06327 60 57.06352 48.54127 35.26748 18.54146 0.000478 18.5406 35.2667 48.5407 57.0632 - 60 57.0636 48.5414 35.2676 18.5416 0.00064 Preac=Pimag 0 6.180335 11.7557 16.18033 19.02112 20 19.02114 16.18036 11.75574 6.180385 5.31E-05 6.18028 11.7557 16.1803 19.0211 - 20 19.0212 16.1804 11.7558 6.18044 0.00011 6.180234 11.75561 16.18027 19.02109 20 19.02117 16.18042 11.75583 6.180486 0.000159 6.18018 11.7556 16.1802 19.0211 - 20 19.0212 16.1805 11.7559 6.18054 0.00021 P appparent 0 7.159203 15.57535 24.42461 32.84076 39.99997 45.20145 47.93604 47.93605 45.2015 40.00005 32.84086 24.42472 15.57546 7.159303 7.96E-05 5.20142 7.93602 7.93607 5.20153 0.00011 7.159069 15.5752 24.42446 32.84063 39.99987 45.20138 47.93601 47.93608 45.20157 40.00016 32.841 24.42487 15.57561 7.159437 0.000186 5.20135 - 7.936 7.93609 5.2016 0.00021 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -