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

[@angerwaqas2969]

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]

Good Luck dear.

[@randomniqqa]

Thanks from india 🙏

[@ElectricalEngineeringAcademy]

Welcome dear.

[@shehzadhussain595]

Sir ap k lectures suny k bd mary sary concepts regarding three phase circuits clear hovy .

[@ElectricalEngineeringAcademy]

Thanks Shehzad. Good Luck.

[@musfiqurrahman2982]

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]

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]

@@ElectricalEngineeringAcademy ohhh,so can I use this method in every Y-delta connection for finding line current and phase current?

[@ElectricalEngineeringAcademy]

Yes. Watch this kzbin.info/www/bejne/onebi5Kfr62bbaM

[@aijazmuhammadkhan3139]

Sir parhaaty ap bhi slides se hain par apko slides se parhana aata hai atleast reading nhi krte ye bat achi hai

[@subashacharya2682]

11:53 isn't Ib supposed to be -90 + (-90) = +150 and Ic supposed to be -30

[@ElectricalEngineeringAcademy]

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]

@@ElectricalEngineeringAcademy ty sir

[@kanaktekwani2517]

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]

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]

which book u r following ?

[@kanaktekwani2517]

@@ElectricalEngineeringAcademy thank u so much for this ans🙏🏻

[@aijazmuhammadkhan3139]

@@ElectricalEngineeringAcademy Sir ap ALEXANDER ko follow kr rhe hain na ? Funamental of electric circuits ?

[@Zubairkhan-rb1fx]

nice

[@shehzadhussain595]

Nice

[@amierulhaqimi3428]

i dont need to go to classes anymore wkwkwk

[@sobuzhussain5056]

Sir apni Pakistan ar kothy thaken?

[@ElectricalEngineeringAcademy]

আপনি কোন বিশ্ববিদ্যালয়ে পড়ছেন

[@sobuzhussain5056]

Bangladesh ar Green university ta.

[@sobuzhussain5056]

Sir apni kon university ta job koren? Janta isca korca. Jodi aktu bolten.

[@ElectricalEngineeringAcademy]

@@sobuzhussain5056 UET Lahore

[@sobuzhussain5056]

@@ElectricalEngineeringAcademy Allah jodi amaka kono din sujoke day, tha hola ami akdin apnar sata dekha korbo inshallah.