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RMS VALUE OF NON SINUSOIDAL WAVE — Transcript

by Lizandro Bitang · 6,535 words · 1,138 segments · language en · Watch on YouTube

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0:01Good day everyone. So welcome to our to
0:05the continuation of our
0:08discussion on the topic
0:11nonosoidal waves. So I really now
0:15excited to guide you through the topic
0:19that plays a vital role in modern
0:23electrical
0:24engineering. So this is the RMS or the
0:27effective value of the sinusoidal wave
0:31form. So in this
0:33presentation we go beyond the typical
0:35side wave and explore how to deal with
0:38real world signals which are of
0:41irregular and complex in shape. So let's
0:46begin.
0:52So in electrical engineering the
0:55effective value or the RMS or root
0:58square is used to express an alternating
1:03quantity in terms of its equivalent DC
1:08value particularly in how it delivers
1:11power to a resistive load while senosal
1:17wave form are often used in theory and
1:20textbooks. Most practical signals like
1:23those found in power electronics and
1:26digital systems are nonsenosidal.
1:30So for example we have this square waves
1:34triangular
1:36waves and complex distorted signals.
1:40That is why understanding how to compute
1:45RMS values for such wave forms is both a
1:48theoretical necessity and a practical
1:55skill. So let's define it formally. The
1:59root means square value of a periodic
2:01wave form is equivalent to the square
2:04root of the
2:06average of the squares of the
2:08instantaneous values over one cycle. So
2:13we will just we will use this formula
2:18no. So why is the RMS
2:24value? This is because it reflects the
2:28true power capability of an AC source.
2:33In resistive
2:34circuits, it determines how much thermal
2:38energy will be generated.
2:40Whether we analying power supplies,
2:44motors and communication signals, RMS
2:48provides a consistent and comparable
2:52measure of
2:54energy. In essence, it gives us a fair
2:58comparison between AC and
3:02DC power.
3:09So now let's explore the core of our
3:12topic. How to calculate RMS for
3:17nonsenosoidal wave form. Unlike pure
3:20sinusoids, nonsinosoidal signals may
3:23contain multiple harmonics or follow
3:26shapes like square or triangular
3:30patterns. So for example we have the
3:33square wave. So to compute for the
3:35amplitude or I mean for the RMS value of
3:40this wave. So we have here the piece
3:44wise function. So for
3:47example you have this square wave no
3:51from 0 up to time t for one full cycle
3:56or one period. So labeled as T sub O. So
4:01half of it will be equivalent to t / 2
4:06and the other half is t / 2. So for the
4:10first function from
4:130 to t / 2 you have this a no equation a
4:20from 0 to t / 2 and from t / 2 to t so
4:27you have this -
4:30a just substitute this to our
4:34formula okay so for the full cycle for
4:37one cycle of the wave
4:39have this the positive cycle plus the
4:42negative RMS of the negative cycle. So
4:46after manipulating the
4:48integral so come up with the value of
4:51the RMS of a square wave equivalent to
4:56its
4:58amplitude. Next.
5:02So for a triangular wave having a pick
5:04value of a so RMS value is equivalent to
5:08the amplitude divided by s root of 3. So
5:14the derivation of this formula will be
5:17left as your
5:20exercise. And also for a soot wave for a
5:24pick value of a
5:26again so we have a value of a divided by
5:30s root of 3. So take note of this
5:35quantities or
5:40equations. So for a general complex wave
5:43so the instantaneous current as a for
5:45example this instantaneous current as a
5:48function of time. So this is composed of
5:52several harmonics no so they have
5:55fundamental second third up to and
5:59harmonic so a sub0er here represents the
6:02dc component so im1 im2 im3 imn are the
6:09maximum values of each harmonic and
6:14alpha 2 alpha 3 are the face angl
6:20and omega is our angular frequency and t
6:24represents the time.
6:28So let's solve for or let's find a
6:32formula in solving for the RMS value of
6:37this complex wave.
6:41So we will just use the formula for
6:46solving the RMS value. So we just
6:49substitute the instantaneous value or
6:52the instantaneous
6:53equation. So you have this quantity
6:57square dt from 0 to t for one period or
7:03for one whole cycle.
7:05So after manipulating this integral so
7:09come up with this effective value or RMS
7:12value equivalent
7:13to square root of the DC component squ
7:16plus the summation of the individual
7:19maximum value on each harmonic and
7:23squares so divided by
7:262 and also we can express that equation
7:30in terms of RMS value of each harmony.
7:35Okay.
7:38So uh take note that RMS is equivalent
7:41to the maximum value divided by SO 2 for
7:45aign wave. Ha? This is just applicable
7:47for a sign
7:49wave. So I over s root of 2. So
7:55substituting this value to the equation
7:58previous equation. So we arrive on
8:02this uh formula for RMS value in terms
8:06of the RMS value of each
8:10harmonic. Okay. So we can also use this
8:13formula if you are solving for harmonic
8:16of each uh RMS value of harmonic so you
8:19can use this formula.
8:26And also we can also solve for the power
8:30no power due to
8:32anosidal voltage so we have here the
8:35general expression for the average
8:37power in a circuit with time varying
8:41voltage and current given as this
8:43equation.
8:45So I already introduced this equation to
8:48you in our previous
8:50discussions. 1 / t integral of e of t or
8:55the instantaneous value of voltage times
8:57instantaneous value of current dt over 1
9:01period from 0 to t.
9:05So for example if you have
9:08this nonsusidal voltage oranous voltage
9:12equivalent is em sub 1 sin omega t +
9:16alpha 1 + em sub 2 sin 2 omeg t so and
9:19so forth and for
9:22current im sub 1 sin omega t + alpha 1
9:26prime 1. So we just substitute
9:30this uh equations to our formula for
9:36power then integrate it from 0 to t for
9:40one period. So after integrating the
9:44equation or the
9:47integral so we come up with this formula
9:51for power. So is this is this is just a
9:55summation of uh its maximum product of
10:00maximum value of each harmony divided by
10:042 and cosine of the difference of the
10:07two phase
10:09angles. So em sub 1 and im sub one
10:12corresponds to the fundamental harmonic
10:14and you will also use the uh face angle
10:18of its harmonic of fundamental harmonic.
10:22So for voltage you have alpha 1 and for
10:25current alpha prime
10:281. Every everything will just
10:32follow. So take note of the
10:39formula. So to to better understand our
10:43formula. So let's apply it to our
10:46example. So find the power represented
10:49by the following
10:52given. So instantaneous voltage
10:54equivalent to 100 sin omega t + 30 - 50
11:00sin 3 om t + 60. So 25 + 5 sin 5 omeg t
11:08and for current 20 sin omega t - 30 + 15
11:13sin 3 om t + 30 + 10 cos 5 omeg t - 60
11:19amp. So we have now this
11:22two uh instantaneous values of voltage
11:26and current. So for the voltage it is
11:30expressed in terms of a sign function
11:33and for current
11:36so expressed in sign and the fifth
11:40harmonic is expressed in terms of
11:44cosine. So before solving or before
11:47applying the formula see to it that the
11:50quantities
11:51are consistent. So meaning if in the
11:55fifth
11:56harmonic you are using sign of course in
11:59the fifth harmonic of the instantaneous
12:03current it should also be sign so we
12:08convert this function or this equation
12:10in terms of
12:13sin so recall the formula that
12:19cosθ is equivalent
12:22to sin
12:26ofθa +
12:3190°. Okay. So
12:36substitute sin
12:38of 5 omeg
12:43t -
12:4660 + 90.
12:51So this quantity 10 cos 5 omeg t- 60° is
12:55equivalent to
12:5810 sin
13:02of
13:045 omega
13:10t
13:13plus 30.
13:15Okay. So this this is the equivalent sin
13:19of this cosine quantity. So we can now
13:23use the formula.
13:27Okay. So, so for for the fundamental
13:30harmonic so the maximum value for
13:33voltage is 100 times the maximum value
13:36for current 20/ 2 cos of the face angle
13:41of the voltage which is 30
13:45minus cos uh minus the face angle of the
13:48current which is nega 30.
13:53Next for in the third
13:56harmonic so will have
13:59-50
14:0115/ 2 cos of the face angle of the
14:06voltage the third harmonic which is
14:0960 min the face angle of
14:15the current the harmonic 30 and in theth
14:20harmonic
14:22So we have this maximum value of
14:2525
14:2710/ 2 cos of the face angle of the fifth
14:32harmonic is zer so 0 min the face angle
14:38of the third harmonic which
14:40is I mean in the fth harmonic which is
14:4430
14:46so after substituting the vales
14:50No. So you can now solve for by
14:54using your calculators. You can solve
14:57for current or the power I
15:00mean.
15:01Okay. So for this quantity you have this
15:06500 watts. And for the next quantity for
15:11the third harmonic you have - 324.75 75
15:15and for the fifth harmonic you will have
15:18108.25
15:2025. So summing up all of this will
15:25result to the power which is equivalent
15:29to
15:33283.5. Okay. So let's proceed to the
15:40next. So we have here another example.
15:44So find the power delivered by the
15:48following instantaneous values of
15:51voltage and current.
15:54So you have here for the voltage you
15:56have a value 100 sin omega t + 50 sin 5
16:01omeg t- 80- 40 cos and for the current
16:07we have 30 sin of omeg t +
16:1160 sin of 5 omeg t - 50 and sin 7 10 sin
16:177 omeg t + 60 so as you can observe that
16:23the seventh harmonic for the voltage
16:25source is expressed in terms of cosine.
16:30So we convert this into sin. Okay. So s
16:36function so step one we have here
16:39convert all to sign function. So the -
16:4240 cos 7 om t + 30 is equivalent to in
16:47terms of s this is equivalent to
16:50-47 omeg t +
16:5520°. So solving for the
16:58power same as what we did on the
17:00previous
17:02example solve for the
17:05individual power on
17:07its harmonic. So in the fundamental
17:11harmonic you have here 100
17:1530/ 2 cosine of so your face angle for
17:20the fundamental harmonic of your voltage
17:23is
17:24zero and for your current is 60 so 0-
17:2960 so using using your
17:33calculator so you will come up of this
17:36on this value of 750 50 wat and next in
17:41the fifth harmonic you have
17:44433 and in the 7th harmonic you
17:47have 100 so take the sum of this three
17:52powers no so you come up with this value
17:58of value of power which is equivalent
18:08So ah
18:13okay this value of
18:22sorry
18:27okay so volt ampires or your
18:34uh apparent power and power factor for
18:38nonosidal waves. So the volt amper
18:41apparent for the apparent power. So volt
18:44amper so this is determined by the
18:46product of the effective voltage and the
18:48effective current.
18:50So you are already familiar with
18:54it that S or the apparent power is
18:57equivalent
18:59to the voltage times the current SBI. So
19:05this is in
19:08terms for the volt
19:10amper we just use this formula. So take
19:15the RMS value of the voltage no and also
19:19take the RMS value of the
19:21current and multiply
19:25it. So for
19:27example so in our previous
19:29example you have here the
19:33the RMS value for
19:36voltage and RMS value for current. So
19:39just take the RMS value of each
19:41quantity.
19:43Sige. So, so you have here the value
19:49for it's a parent power equivalent to
19:5510001
19:56vol. So just take
19:59the uh RMS value of the voltage and the
20:03RMS value of the current and multiply
20:07them will have the amp the apparent
20:12power.
20:14Next. So for the power factor so power
20:18factor this is the ratio
20:23between ratio between the real power and
20:26the apparent power. So this is a measure
20:30of how much real power has been consumed
20:34for the given apparent power. Okay. So
20:39you just substitute the formula. So this
20:41is the formula for the real power p no
20:46real power p
20:49and the denominator is the apparent
20:58power so this is p or
21:03div va
21:06or vol ampes
21:12So for example in our previous example
21:15in example 6 so we compute we computed
21:19this equivalent wats no power and in our
21:23prev example 7 so we computed
21:27this volt amp the apparent so just
21:31divide
21:32p
21:34va come up with this with this this
21:38value
21:39of our
21:43factor. So na siya no ratio between the
21:49real power and the apparent
21:56power. Okay. So let's now come to
21:59circuit analysis when waves are
22:02nonsenosoidal.
22:04So we just walk through on this example
22:07to better understand how we're going to
22:09analyze a
22:11circuit when our source source voltage
22:16and current are expressed in
22:21nonosoidal. So for
22:23example given the circuit with the
22:26parameter shown
22:30so these are the parameters of our
22:33circuit. So a series
22:35circuit no so when omega or the angular
22:38frequency is 377 radians per second and
22:43the voltage source is
22:45141.4 sin omega t + 70.7 sin 3 om t +
22:5130° - 28.28 28 sin 5 omeg t-
22:5620° volt is impressed.
23:00find the current I so the reading if you
23:04have if you connect here an
23:08ameter that ameter would read and also
23:11find the total power dissipated and the
23:15effective value of voltage drop across
23:18the inductance and also find the
23:21equation for the current wave so to
23:27solve this problem no uh we will
23:30consider the current on each fonda on
23:36each on each harmonic. So for let's
23:39first analyze using the fundamental
23:41harmonic and second let's analyze the
23:45current or the voltage drop in the third
23:49harmonic and also in the fifth harmonic.
23:54So first let's do the analysis in the
23:58fundamental harmonic. So we have the
24:00given. So you have R equ to 6 ohms L
24:050.05 Henry and 98.8
24:09microfarad. Then the source voltage is
24:13given. So what we are going to solve is
24:17the current
24:18I the total power dissipated effective
24:21value for voltage drop across the
24:24inductor and in equation of the current
24:28wave. So take note that the ameter would
24:30only read
24:32RMS value ha the effective
24:35value. That's what our ameter would
24:39read. So first let's analyze the circuit
24:43in the
24:45fundamental harmonic. Okay. So
24:49fundamental
24:50harmonic. So you have your voltage
24:52number one equivalent
24:55to so in the fundamental harmonic. So
24:58you have this value
25:00of your voltage maximum
25:04141.4. So ah let's solve for its RMS
25:09value. So we will divide it by root of 2
25:14since this is expressed in sign. So
25:16141.4
25:174
25:192 100
25:23volt inductance in the fundamental
25:25harmonic so we will use the omega
25:28equivalent to
25:32377 this is because in the fundamental
25:36harmonic you have your angular frequency
25:38as
25:41omega okay so 377 times the inductance
25:45you have this
25:4718.85 ohms and for your capacitance or
25:51capacitive reactance again use
25:55the
25:57value of omega to
26:00377 so you have this 26.85 85 ohms and
26:05the impedance on the fundamental
26:07harmonic this will be equivalent to 6 +
26:11j xl-
26:14xc so you have this impedance 1
26:18equivalent to 6- g8 so meaning in the
26:22fundamental harmonic our circuit behaves
26:25as like a capacitor or rc so we can uh
26:32we can We now have an idea that in the
26:36fundamental harmonic our current is
26:39leading the voltage. No, since this is
26:43an RC it behaves like a capacit capacite
26:48RC circuit no so expressing this or
26:53taking
26:54the uh magnitude of
26:57this quantity so you have this
27:00equivalent to 10 ohms. So solving
27:04for the RMS value of current no in the
27:09fundamental harmonic so we will use the
27:11RMS value of voltage 100 / by the
27:15impedance no 10 so uh we have a value of
27:2110 amp so for
27:23power so we can use the formula I squ
27:27only the resistance will consume the
27:30real power
27:31watts so Use the current since this is a
27:34series circuit
27:37I s R so 100
27:42so we have a power in the fundamental
27:46harmonic equivalent to 600 watts and the
27:50voltage drop across the inductor in the
27:53fundamental
27:54harmonic so I times the inductive
27:59reactance which is 18.85
28:0285 so you will have 188.5
28:055 volts and the face angle no solving
28:10for the face anglea 1 so you will have r
28:16tangent so we just use the impedance
28:21triangle or we are solving the power
28:24factor on the fundamental harmonic so
28:28the resistance is
28:306 and inductive uh capacitive reactance
28:35of J8 -
28:38J8 so you will have
28:41this angle of impeda angle equivalent to
28:46- 53.1°
28:4912° so meaning in the fundamental
28:53harmony you will have the expression for
28:55current let us say
28:58I1 this will be equivalent
29:02to the RMS value of
29:05current this is equivalent to I mean the
29:08maximum 10
29:11s ro
29:132 sin
29:19of omega t.
29:22Okay. So let's analyze. No. As you can
29:26observe
29:27in our voltage source, our voltage V has
29:32a face angle of
29:34zero. Okay. Face angle of zero. So I
29:38will just write it here. So if we're
29:41going to plot it in
29:43our cartesian plane, so your voltage is
29:48located just in the origin V.
29:52No. So V maximum or the
29:57VRMS. Next your
30:01current is at angle 53. Since this is
30:07our circuit the fundamental harmonic
30:09behaves as an RC like an RC. So meaning
30:13current naglead siya sa voltage by this
30:17angle
30:1953.1. So meaning your current ay naad
30:24rin
30:25naka-allocate. So angle between your
30:27current and voltage
30:29is
30:3453
30:3512 degrees. So to express this current
30:41no to express the sinusidal equation for
30:43this current so the location of this
30:46current from the origin is
30:48at+
30:5153.32 so ito siya express as
30:5953.12° so this is your expression for
31:02your eye in the
31:04fundamental harmonic. So 10 ro 2 or just
31:09I is im sin omega t +
31:1453. 12. Okay. So we are not finished
31:18yet. So let's proceed to this next
31:20harmonic. So in our voltage source so
31:23the next harmonic is the third harmonic
31:26having a maximum value of
31:28voltage equivalent to 70.7.
31:33So to take the RMS value of the
31:36voltage. So we just divide the maximum
31:40by s root of
31:432 and our inductive reactance of course
31:47this will change because the angular
31:49frequency is changing. So our angular
31:54frequency now is 3 times the fundamental
31:59frequency. So we just multiply the
32:03XL1 3 at times 3 anghang omega. So
32:09therefore we have now our XL3 equivalent
32:12to 56.55 55 and also the uh the
32:18capacitive reactance no so if going to
32:22change the increase
32:24the uh value of angular frequency so our
32:29inductive capacitive reactance
32:32will be will decrease no so will
32:37decrease by the multiplying factor no if
32:42we are if we are multiplying the or
32:45increasing
32:46the uh angular frequency by a factor of
32:503 our induct in our capacitive reactance
32:53will decrease by 3 or divided by 3 so
32:59therefore our xc will be equivalent to
33:028.95 95 so kung nagliog
33:07moon
33:09so that xc is 1 /
33:152π
33:17fc or omega c 1 / omega
33:22c mas dali
33:261 / omega c and if we triple this omega
33:34No so making c as our c will not change
33:38or our capacitance will not change so
33:42the value of xc will just be divided by
33:453 or if we double
33:48the so if we double the value of our
33:52angular frequency so our our xc will
33:56divided by
33:572 okay
34:00so that is how you compute for the XC on
34:06this the next
34:07harmonics. Okay.
34:10So at
34:13solbonia impedance so impedance in the
34:16third harmonic will not be and our
34:19resistance will not change
34:22ha constant no because R is not a
34:26function
34:27of any quantity. So R is the function of
34:32uh the angular frequency. No so that is
34:37why even if uh in the fifth harmony
34:40fourth harmonic so I will our R will
34:44remains the
34:46same. Okay so substitute 6 + JXL - XC so
34:526 + J 47.6 6
34:57ohms. So in the third
35:00harmonic you can see that our impedance
35:05function behaves like an RL circuit. So
35:11we already have an idea that in the
35:14third harmonic our current eye now logs
35:18the voltage by a certain angle no
35:22logging naong current danhe. So at timan
35:26ha so taking the magnitude of this
35:30quantity so you will have this
35:3248.1 ohms and solving for the RMS value
35:36of current in the third harmonic.
35:39So voltage RMS div
35:42by and that magnitude of the impedance
35:47so na siya
35:511.04. Now solving for power in the third
35:55harmonic. So we will use the third
35:59harmonic current square times the value
36:02of resistance R. So naatay 6.48.
36:0848
36:10wats the third harmonic and also we can
36:13solve now for the voltage drop across
36:16the inductor in the third
36:19harmonic so we have 58.9 9 vol and
36:26solving also for the face angle or the
36:28factor
36:29angle so we will use the power or I mean
36:34the impedence triangle
36:38so for our impedence
36:42triangle so
36:44the real part will be the r which is 6
36:49and the inductive part or the reactive
36:53part will be
36:5856
37:0055. Okay. So taking the R tangent
37:04opposite over adjacent for the XL / R.
37:09So we have this R tangent equivalent
37:13to R tangent of
37:1547.6/ 6 equ to 82.8.
37:20So what does this
37:23imply? So same as what we did in the
37:26fundamental harmonic. No. Okay. I will
37:29erase
37:34this. Okay. So as you can
37:38see our voltage no if we look back in
37:43the previous slide. So our voltage B is
37:48at 30.
37:52de socated pos
37:5830 so meaning our voltage
38:02b let us
38:06say 30°
38:09from the
38:13origin and since in the third harmonic
38:17our current eye is logging the voltage
38:20by certain angle and that angle is this
38:23angle so our eye is located somewhere
38:31here the angle between our voltage and
38:33current is
38:40828 so therefore the distance of this
38:43current I from the origin will now be
38:45equivalent to 82.8-
38:498- 30. So this is equivalent
38:53to uh
38:5982.8
39:02828 - 30 so we
39:06have 52.8.
39:11So we can express
39:14the
39:16uh current in the third harmonic no so I
39:22sub
39:253. So this will be equivalent to so RMS
39:29s of 2. So 1.04
39:3304 ro
39:362 sin
39:38of so since this is in the third
39:41harmonic o third harmonic sa na omega t
39:46and kin siya naman siya paubos no
39:52pa-counterclockwise I mean pa-clockwise
39:55clockwise direction so meaning the angle
39:57is negative so-
40:03528°. So this is your expression for
40:06current in the third harmonic. So 1.04
40:10ro 2 sin of 3 omeg t -
40:1452.8. So we have we have the last
40:18harmonic in the fifth
40:21harmonic. So fifth harmonic.
40:25So in the fifth harmonic we have the
40:26current or our voltage equivalent to
40:2928.
40:3128. I will look back. So
40:3628.28 at
40:40-2. So 28.28. So solving for the RMS. So
40:45we divide it by s of 2 and our inactive
40:49reactance will multiply by 5 no
40:54multiplied by 5 and our capacitive
40:58reactance will be divided by 5 our
41:02fundamental capacitive reactance will be
41:04divided by 5 and our
41:07fundamental uh inductive reactance will
41:11be multiplied by
41:135 So solving for the impedance in the
41:17fifth harmonic. So naay 6 + JXL which is
41:2494.25- capacitive reactance which is
41:275.37 so we have 6 + J 88.88 88 ohms. So
41:34as you can observe that impedance in the
41:37fifth harmonic behaves like an RL
41:40circuit since this is positive so RL
41:44siya then expected na that it is
41:48expected that our current in the fifth
41:50harmonic logs the voltage by a certain
41:56angle since RL circuit the R circuit
42:01ising the voltage
42:03So taking the magnitude of this quantity
42:07have this 89
42:09ohms and solving for the current in the
42:12fifth
42:13harmonic. So just substitute we have now
42:16the voltage RMS div the impedance. So
42:21you will have RMS carmonic equivalent to
42:260.225.
42:30So solving for the
42:32power. So this is equivalent to the
42:35square of the current times the
42:38resistance. So therefore you will have
42:41you have now the value for power in the
42:43harmonic equ
42:46to304
42:48watts and also the voltage drop across
42:53the inductor in the fifth
42:56harmonic you will have current times XL.
43:00This is equivalent
43:02to 2 and for the face
43:07angle so for the face angle so the same
43:11so tangent of XL over R which is
43:16equivalent to 86.1°
43:191 de okay so solving for the total RMS
43:25current or the current that our ameter
43:28will be reading this will be equivalent
43:32to the square root
43:34of the RMS in the
43:38fundamental square plus RMS value in the
43:42third harmonic plus the RMS value in the
43:46fifth harmonic so namay mga y so just
43:50substitute 10 s + 1.04 s + 20
43:550.225 square so you will have the RMS
44:00value or the total current this will be
44:03equivalent to 10.05
44:0705 and for the power so we just sum up
44:11no all the power consumed on each
44:16harmonic so the first fundamental you
44:19have 600 the third harmonic you have 648
44:236.48 48 and the third harmonic uh we
44:27have a value of
44:2934304 so in all we have total power
44:34what's consumed is
44:37606 8 and of course the total inductive
44:42voltage in the inductor so same lang
44:46kuha gap on to RMS sa VL so this will be
44:49equivalent to
44:51VRMS VLMS S in the fundamental, VLRMS in
44:56the third and VLRMS in the fifth
45:01harmonic. So ay
45:04198.8
45:06volts. Now for the equation of current
45:09wave form no at siyang-create sa una.
45:14So for the current in the first
45:16fundamental harmonic so
45:20naay
45:22pare
45:24114.14 sin omega t + 53.2 So this is the
45:29equation no equation where current in
45:32the fundamental harmonic. In the third
45:34harmonic we have this
45:371.47 sin of 3 om t -
45:4352.8. And in the third
45:45harmonic I mean in the fifth
45:47harmonic so you'll have
45:50this root of 2 * 0.25 sin of 5 om t -
45:56106.1 1 equivalent to
46:01-0.318 sin of 5 omeg t- 100.1
46:10one negative siya
46:15soat so take note that our
46:19voltage has
46:22a maximum value of negative
46:2628 and located at -
46:3220°
46:35so dinis
46:40ne 20 or
47:0020
47:02and and since
47:04our circuit in the fifth harmonic
47:08behaves like an inductive RL
47:11circuit. So logging ang kurente. So the
47:15angle between your current and voltage
47:19will be equivalent to 86.1.
47:23So somewhere
47:27here so this will be equivalent
47:31to
47:3586
47:371 so your voltage here is ne
47:4728.28 Ayan 28 no? Ano ba
47:52'to? -
47:5528.
47:5628. So if you have this voltage negative
48:00of course your current will also be
48:01negative so negative
48:04ay sub
48:065. So this angle here is
48:1186.1 and the distance of your i5 from
48:14the origin will now be equivalent to
48:1686.1
48:1920. So the expression of your current
48:22will be equivalent to this
48:27ne 2 the maximum current which is
48:330.25 sin of 5 omeg t - 106 because 86.1
48:38+ 20 is
48:40106.1 so negative si ha so - 106.1
48:45or
48:48-0.318 sin 5 om t +
48:53106.1 and therefore the complete
48:56expression for your current will now be
48:58equivalent to I1 I3 +
49:02I5 so morning naasay one
49:07ah
49:1014.4 sin of 53.12 12 +
49:14247 sin of 3 omeg t -
49:1752.8 - sin 5 om t -
49:23106.1 or we can also express this
49:28quantity in the other direction and it's
49:31opposite direction.
49:33If we going to take the opposite of this
49:35I so location will now be here. So if
49:39this is
49:40negative
49:42vector opposite vector no so this act in
49:47this direction and its opposite will be
49:49negative of
49:50this
49:52direction. Yeah. If this is 106.1
49:58And this angle now will
50:00be so an so since this is 106.1 so kay
50:0890 so 106.1-
50:1190 angle and this angle is just equal to
50:14this
50:16angle this is theta of course this angle
50:20is theta so 106.1
50:241-
50:2590 this is equivalent
50:31to 16.1 no
50:3616.1 and this is 16.1
50:40also. So 90- 16.1 so makuha so
50:4990- 16.1
50:51So have
50:5473.9
50:57[Musika]
51:0073.9
51:0273.9° so that is why we also have here
51:04another expression equivalent to and
51:07positive na ha I is positive so pos
51:120.318 318 plus ano positive
51:17+318 sin of 5 omeg t + 73.9 9 amp. So
51:22this is the complete expression for the
51:25current or you can also you can can have
51:27your answer as
51:30kanilang
51:33no pwede
51:36na
51:38kanya or pwede po kini i answer si ha.
51:44Okay. So, okay. So, so let's proceed to
51:47another
51:49example. So, this is we will now be
51:52having an example for a parallel. So,
51:55that's for series in the
51:57previous sa parallel. So, for example,
52:00given a circuit shown in figure 23 with
52:0460 cycle constant as
52:06shown 60 cycle 60 herz. So when the
52:10voltage
52:13V when the voltage V is equivalent to
52:16141.4 sin omega t + 7 sin omega t in the
52:21fundamental
52:23harmonic
52:25and in the third
52:27harmonic naatay
52:3170.7 plus 30 it's face angle is 30 and
52:34the third harmonic I mean in the fifth
52:36harmonic - 28.28 28 sin 5 om t -
52:4120 find the ameter value of the total
52:45current i
52:48so the reading of our ameter the current
52:52in each branch power dissipated by each
52:56branch total power dissipated and the
52:59equation of the resultant
53:01current so our omega or our angular
53:04frequency will be equivalent to 377 7
53:08radians per seconds. So this is our
53:12circuit no parallel branches. So RC 5
53:17ohms and 15 ohms. 5 ohms ang R C is 15.
53:22And R in the second branch connected by
53:26CD point CD. So 10 ohm resistance and 2
53:30ohms inductive reactance. Okay. So let's
53:34now solve for
53:36the um anouns nouns quantity. So as what
53:42the same as what we did on the previous
53:45example so in solving for a parallel
53:49same procedure lang. Soccerons ang iyang
53:53mga components. So for first solve for
53:56the fundamental component second in the
53:59third component or whatever component is
54:02present.
54:03And the last is the last third component
54:06or fifth fifth harmonic. Soon
54:10lang depende na kung voltage source or
54:14even current source kung upat siya na si
54:16harmonic 7 so kaupat po ka or kalima po
54:21ka mo mag-solve consider good mo kada
54:24component or kada harmonic. So first
54:27move si ka sa fundamental.
54:31So in the fundamental
54:33harmonic angipangita niya is ang current
54:37voltage sa power no total
54:42power so in the fundamental harmonic our
54:45voltage or our maximum voltage is 141.4
54:504 so divide siya 2 para makuha na to ang
54:55iyahang ah iyahang RMS value. So 100 +
55:03J0 ang voltage
55:07sa RM voltage sa fundamental. So current
55:11in the first branch can see in the first
55:19branch so to solve for the current in
55:21the first branch we need the impedance
55:23in the first branch so namay voltage
55:27100/ impedance ng 5- j15 so in the
55:30fundamental harmonic mam itong value
55:33na
55:35j15 or - j15 minus kay capacitor 1. So
55:41in the first branch we have a current
55:44of 2 + J0 or 2
55:50amp. So currents in the second
55:54branch or in the CD branch
55:58kini. So voltage divided by impedance
56:0110. So iyung quantity iyang value sa
56:03impedance is 10 + J2.
56:07So that is why
56:08100/ 10 + J2. So result niya is
56:139.62- J
56:161.925. So taking the magnitude of this
56:19current so naay
56:239.82
56:25amp. So next atong ang total in the
56:30fundamental.
56:32So total niya ang total current is
56:35just summation of the current on each
56:39branch
56:40parallel current plus current current in
56:43the fundamental harmonic. So that is
56:47IAB1
56:48plus
56:50ICD
56:521.
56:54Okay. So our current therefore will be
56:57equivalent to this quantity 12.323
57:0333.
57:06Okay. Next. Solve for the face
57:11angle. So how to solve for the face
57:14angle? So siya. So at kuha iyahang
57:19equivalent
57:21ng pwede lang gamit an current no. So
57:25real
57:29uh current the real component is
57:31equivalent to
57:3611.62 iyahang imaginary component kanina
57:40J
57:41ngang
57:454.075 so nag tangent ha r tangent of
57:494.075 / 911.62
57:5362 so naay angle equivalent to
57:5919
58:014 okay so next solve for the power in
58:05the first branch so naman tay current
58:09diha so
58:13power this will be equivalent
58:18to okay real naman eh wajer component so
58:21pwede lang anim ah the real part times
58:25the voltage so natay 200 watts and also
58:30in the second branch CD branch so 100
58:34voltage times 9.62
58:3662 real component. So pwede lang na
58:40gamitan 962 wats
58:44ng ah power in
58:48the second branch on the
58:51fundamental harmonic.
58:54Sunod and the third harmonic. So third
58:58harmonic our na our frequency or angular
59:03frequency 3 no so mahitabo ang
59:09XC maivide by 3 while our inductance or
59:15our inductive reactance XL ma-multiply
59:18by 3. So RMS value in our kuan sa volt
59:25in the third harmonic so naay 50 volts
59:30or express in kuan in rectangular
59:33coordinates of 50 + J0 imaginary
59:38component and solving for the current in
59:40the AB branch in the third harmonic soay
59:4550 volts div 5-
59:49j5 so 5 +
59:52J 5 so magnitude nga 7.07
1:00:0007 so in the CD branch 10 +
1:00:04J at 50/ 10 + J6 so
1:00:09na value 3.68 - J
1:00:152.21 in the so iang magnitude naay 4.33
1:00:20so atong i-adding du ha vector addition
1:00:23ha so 5 + 3.68 68 naay
1:00:278.68 and 5
1:00:31J5 - J2.21 so naay 2.79 so iyahang
1:00:36magnitude
1:00:399.11 so kung ano to
1:00:42i-angle so kani lang ah the imaginary
1:00:45component divided by the real component
1:00:48soay 17.85
1:00:5385 de so the same to solve for the power
1:00:58gamit real
1:01:00component 5 voltage
1:01:0550 then voltage ng 50 times sa real
1:01:08components ng current ng branch 3.68 68
1:01:14so ay
1:01:16184 watts so muna siya for the third
1:01:21harmonic next last harmonic fifth
1:01:24harmonic so nahimong times 5 ang omega
1:01:30no soong current or
1:01:37voltage so
1:01:39rms 20 or
1:01:43orang
1:01:45voltage equivalent to in terms of ah in
1:01:50terms of tangular 20 +
1:01:54j and
1:01:56karon ayahan ng ayaahan naing kuan
1:01:59ayahan naing xc ma-divide by 5 gikan si
1:02:04ang fundamental ang fundamental is 15
1:02:08man 5 so thatay Sorry ohms ah nahiya
1:02:13hading ko an indu uh inductive inductive
1:02:18reactance ma-multiplied by 5 so
1:02:23naay 10 ohms so the same procedure solve
1:02:28ka sa current in the first branch so RMS
1:02:32voltage divided by current naatay ken
1:02:352.94 + J.76 76
1:02:39763 then taking its magnitude you have
1:02:43this 3.43 amp and the next branch CD
1:02:47branch so again 20/ 10 + J10 so ayusan
1:02:54mo i-mine ang negative
1:02:56haon ang k lang nagdetermin n direction
1:03:00ha ayaw siya na i-mind only magnitude
1:03:02sa so naatay 1- j1 or 1.414
1:03:07414 i-add na to ang duha ka current king
1:03:11mga current din i-add para makuha na to
1:03:13diha in the fifth sa fifth
1:03:15harmonic so
1:03:17naatay current nga 3.94 +
1:03:23j.63 or
1:03:254.01 amp so solving for its angle tayo
1:03:3010.95
1:03:3395° so solving for the
1:03:36power so voltage times the real
1:03:39component of
1:03:41current so 58 and 20 plus 1 only 20 wats
1:03:49so sunod ito na kuha
1:03:52total so the amity reading is ma ang
1:03:55total current at point f so total
1:03:59current is the fundamental
1:04:0112 point or it's RMS value
1:04:0412.33 plus the RMS value of current
1:04:08in the third harmonic square plus the
1:04:11third or the RMS value of current in the
1:04:16fifth harmonics. All in all naay 15.9
1:04:20ang maasa sa atong ameter so sa kada
1:04:24branch sad so at
1:04:27gamitan square root of RMS value of
1:04:30current in the first
1:04:32branch plus the square of the arm's
1:04:36value of current in the second branch
1:04:37plus the arm's value of current in the
1:04:39third branch square. So naay 10.1 one
1:04:42and last in the CD
1:04:45branch mga values
1:04:47balik so
1:04:50value so sa first
1:04:54branch first
1:04:56branch first
1:05:02branch
1:05:10Okay. So the total power ato lang
1:05:13ipang-sum tanan in the first branch plus
1:05:18tan i-add na to sa second branch cd
1:05:21branch i-add tanan. So ang total
1:05:25niya plus power AB plus power CD have
1:05:30this
1:05:32quantity. Now sa current wave form
1:05:36current wave form the fundamental
1:05:39branch
1:05:43so ah balik ha. So in the fundamental
1:05:47branch so diha current kini 12.33
1:05:5233
1:05:54okay 12.33 33 maxim RMS na times na to
1:05:59sa iyang s of 2 para makuha tag im in
1:06:02the first branch or i mean in the
1:06:04fundamental component so sin omega t +
1:06:1019
1:06:1244 so kung imohang kuanon kung imohang
1:06:16solbon ang total impedance sa
1:06:18circuit considering the
1:06:21fundamental
1:06:22component so diha na makita Kung un sa
1:06:27ato ang
1:06:27circuit it could be RL or RC. So let's
1:06:34check para dali lang. So ato ang formula
1:06:37na parallel
1:06:39lang itong parallel or pwede to in terms
1:06:42of admittance no pwede tay in terms of
1:06:46admittance if daghan na siyang branches
1:06:49or pwede sad
1:06:51duhara para ato lang gamitan
1:06:57formula sa parallel ng dua branch lang
1:07:00du ka branches so as you can see that
1:07:05Our
1:07:07equivalent impedance in the fundamental
1:07:10component is equivalent to 7.66 - J
1:07:182.69. Okay. So un na siya so
1:07:22RC no behaves like a capacitive branch.
1:07:29So take note that our voltage in the
1:07:32fundamental na lang siya sa origin
1:07:36zero angle so siya leading
1:07:42current so expected that our i is
1:07:45somewhere above our
1:07:47vle between moto nacompute ito sa i
1:07:53194 so that is why the equation for
1:07:56current in the fundamental branch
1:07:59naay k r value in the
1:08:02fundamental 2 cos of omega t plus the
1:08:06distance sa i from the origin 19.4
1:08:10or siya in the third
1:08:13harmonic o same procedure so kamay
1:08:17bahala mo na mo ah compute check lang
1:08:22check so naay procedure ako
1:08:25nag-introduce previously so just learn
1:08:29from it and also in the fifth harmonic
1:08:33ma so review mo naing siya so finally
1:08:37equation for current
1:08:40So you will now have the equation for
1:08:43current equivalent
1:08:46to ah
1:08:4917.45
1:08:51+ ah sin omega t + 19.4 + 12.9 sin 3
1:08:56omeg t + 47.85 85 +
1:09:035.67 sin 5 omeg 3 +
1:09:08171°. Okay. So that would be all for our
1:09:13discussion. So I hope you learn
1:09:15something and if you are if you have any
1:09:19question so just drop a message on
1:09:24my social media account.
1:09:29Okay. So that concludes our discussion
1:09:33on the RMS value of nonsenosoidal waves.
1:09:37So understanding how to compute and
1:09:39interpret RMS values equip us to work
1:09:43with real world signals.
1:09:46system
1:09:47efficiency and ure safety and accuracy
1:09:51in electrical circen
1:09:55M.

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