Balanced Wye-Delta Connection || Example 12.3 || Practice Problem 12.3 || ENA 12.4 (English)

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Electrical Engineering Academy

Electrical Engineering Academy

Күн бұрын

ENA 12.4 (English) || Example 12.3, Practice Problem 12.3(Alexander & Sadiku)
00:20 - Key Concepts Line & Phase Voltage & Current
01:15 - Steps for solving questions
05:20 - Alternate method of solving questions
06:20 - Example 12.3
09:45 - Practice Problem 12.3
The video explains the concept of Wye-Delta circuit, where source is connected in Wye configuration and load is connected in delta configuration. Concept is emphasized by solving problems.
Example 12.3: A balanced abc-sequence Y-connected source with
is connected to a -connected balanced load per phase. Calculate the phase and line currents.
Practice Problem 12.3: One line voltage of a balanced Y-connected source is Vab =120 angle 20 . If the source is connected to a -delta connected load of 20 angle 40.
find the phase and line currents. Assume the abc sequence.
Fundamentals of Electric Circuits -Alexander & Sadiku
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Electrical Engineering Academy
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Пікірлер: 29
@angerwaqas2969
@angerwaqas2969 5 жыл бұрын
Sir got exam tommorow but this video has made my all concept clear regarding phases. Thumbs up for you keep up the good work!💝
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 5 жыл бұрын
Good Luck dear.
@randomniqqa
@randomniqqa 11 күн бұрын
Thanks from india 🙏
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 10 күн бұрын
Welcome dear.
@shehzadhussain595
@shehzadhussain595 5 жыл бұрын
Sir ap k lectures suny k bd mary sary concepts regarding three phase circuits clear hovy .
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 5 жыл бұрын
Thanks Shehzad. Good Luck.
@aijazmuhammadkhan3139
@aijazmuhammadkhan3139 3 жыл бұрын
Sir parhaaty ap bhi slides se hain par apko slides se parhana aata hai atleast reading nhi krte ye bat achi hai
@musfiqurrahman2982
@musfiqurrahman2982 2 жыл бұрын
why cannot you write that line current =phase current aftrer convertting delta part into wye ? though you are working with wye and finds line current using single phase analysis but for phase current you have used relationship of line current and phase current?
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 2 жыл бұрын
The question is actually asking for phase current of delta load. What we calculated is Phase current (= line current) of Y load (this was done for ease of calculation). So we need to convert I(phase-y) to I(phase-delta).
@musfiqurrahman2982
@musfiqurrahman2982 2 жыл бұрын
@@ElectricalEngineeringAcademy ohhh,so can I use this method in every Y-delta connection for finding line current and phase current?
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 2 жыл бұрын
Yes. Watch this kzbin.info/www/bejne/onebi5Kfr62bbaM
@Zubairkhan-rb1fx
@Zubairkhan-rb1fx 6 жыл бұрын
nice
@kanaktekwani2517
@kanaktekwani2517 4 жыл бұрын
A balanced three-phase star connected source having a line voltage of 208 V rms supplies a three phase balanced delta connected load. The total power delivered to the balanced delta connected load is 1200 W with a power factor of 0.94 lagging. Determine the required load impedance for each phase of delta connected load and resulting line current. Can you help me solve this problem?
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 4 жыл бұрын
from (pf)power factor (=cos theta) calculate theta = 20 degree (approximately). Real power per phase is 1200/3 = 400W. Real power 400=Vline x I line x pf /under root3. so Iline =3.54 A. I phase = Iline/underroot3 = 2.05A. Magnitude of Zdelta = Vphase/Iphase = Vline/I phase = 208/2.05 = 101.46 ohm. Z delta = 101.46
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 4 жыл бұрын
which book u r following ?
@kanaktekwani2517
@kanaktekwani2517 4 жыл бұрын
@@ElectricalEngineeringAcademy thank u so much for this ans🙏🏻
@aijazmuhammadkhan3139
@aijazmuhammadkhan3139 3 жыл бұрын
@@ElectricalEngineeringAcademy Sir ap ALEXANDER ko follow kr rhe hain na ? Funamental of electric circuits ?
@shehzadhussain595
@shehzadhussain595 5 жыл бұрын
Nice
@subashacharya2682
@subashacharya2682 4 жыл бұрын
11:53 isn't Ib supposed to be -90 + (-90) = +150 and Ic supposed to be -30
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 4 жыл бұрын
Yes there is an error in Ib, the angle should be -90 + (-120) = -210 degree. Ic is ok (120 +(-90) = 30), unless you want to use -240 degree option; in that case angle will be -330 degree ( which is equal to +30 (by adding 360 to -330)). Thanks.
@sohaibabbasi3355
@sohaibabbasi3355 2 жыл бұрын
@@ElectricalEngineeringAcademy ty sir
@amierulhaqimi3428
@amierulhaqimi3428 4 жыл бұрын
i dont need to go to classes anymore wkwkwk
@sobuzhussain5056
@sobuzhussain5056 2 жыл бұрын
Sir apni Pakistan ar kothy thaken?
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 2 жыл бұрын
আপনি কোন বিশ্ববিদ্যালয়ে পড়ছেন
@sobuzhussain5056
@sobuzhussain5056 2 жыл бұрын
Bangladesh ar Green university ta.
@sobuzhussain5056
@sobuzhussain5056 2 жыл бұрын
Sir apni kon university ta job koren? Janta isca korca. Jodi aktu bolten.
@ElectricalEngineeringAcademy
@ElectricalEngineeringAcademy 2 жыл бұрын
@@sobuzhussain5056 UET Lahore
@sobuzhussain5056
@sobuzhussain5056 2 жыл бұрын
@@ElectricalEngineeringAcademy Allah jodi amaka kono din sujoke day, tha hola ami akdin apnar sata dekha korbo inshallah.
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