Superposition Theorem || Example 10.6 || Practice Problem 10.6 || ENA 10.4(English)(Alex & Sadiku)

[@OnMyGameplay]

3:25. on the bottom right. We made the voltages short circuit.but why didn t the inductors removed like on the top left one?

[@ElectricalEngineeringAcademy]

Inductor becomes short circuit for dc source. Top diagram has 5V dc source. The other diagrams have ac source.

[@OnMyGameplay]

@@ElectricalEngineeringAcademy but bottom diagram has also dc source ?

[@OnMyGameplay]

Okey i got it thanks

[@NaimurRohman-bk9vl]

Do the two source values need to be the same ? Like both cos or both sine ! What is the problem if you solve the meaning of the source?

[@vishwajeetmalik6486]

Your explaining is so good.

[@ElectricalEngineeringAcademy]

Thanks Malik. Best wishes

[@hellodudes8324]

Sir i am unable to solve the problem at 10:39. It only gives the correct answer if i take the resistor and inductor in parallel but their impedence together seems to be in series

[@ElectricalEngineeringAcademy]

Please whatsapp your working on 923454030919

[@SyedaHamdani-h9j]

Plz tell me how u solve voltage division rule 5:39. How you take 1(5) /total resistance. I didn't get why we take 1 in numerator we dont have value of v1 so how???? Tomorrow is my paperr plz reply me as soon as possible

[@jacstudnet]

V1= total voltage divided by total resitance, multiplied by the resistance across which the voltage is being calculated. So v1 =( (5v /(1+4))× 1)

[@asheertanveer3533]

best video sir

[@ElectricalEngineeringAcademy]

Thanks dear. Best wishes.

[@qaisarzia5920]

thank u sir...

[@bdralm3097]

10:50 this question's answer I'm not getting same answer

[@ElectricalEngineeringAcademy]

please mail me a snapshot at [e-mail]