Відео

As far as I know it is 5/6

@electrical4everyone thank you Professor my test is today I will update you

@@heshamhelal Yes please update me about how did you do and how was the exam compared to previous exams.

@ will do

@@electrical4everyone it was also 5 out of 6 Professor and there were questions about second order transient using laplace

Pout for AC circuit = Vrms*Irms*cos(theta_V - theta_I)

@@electrical4everyone Thank you so much, Mr. Ayman. I just realized that is power factor.

You may add a phase or you may not as far as you have only SIN or COS as the phase is a relative reference. However, if you have both in a question then you should as the 90 phase.

Yes it is 5/6 and I agree with you about your strategy

Sure when you short circuit an independent voltage source, any thing parallel to the source (meaning sharing the two nodes) will be short circuited.

@electrical4everyone thanks Sir, tomorrow is the A1 exam, thank you so much for your super helpful videos.

@binhle8710 good luck in the exam.

when we have second order circuit resonance may happen which lead to increase in output voltage

Please email me and I will send to you. I have A1 tests.

You are welcome

You are welcome

I have silimar Vth answer to yours. My Vth result is 81.3 at 67.49degree.

We cannot open dependent sources. This approach is only correct if we only have independent sources.

@electrical4everyone thanks sir, I got it.

This is from a circuit book

at 17:00 the general form for all 3 cases is given. The same approach applies but with different formula as shown.

I calculated the peak value. if you want the RMS value then you divide by root 2

@@electrical4everyone Ok, thank you

you are welcome

you are welcome

Currently I'm very busy with A1 exam questions. Also, honestly, I'm more comfortable with circuit and power questions. The A5 playlist I prepare it from a previous course I taught.

Are you taking the exam next Monday?

@@tylerbradburn240 yes I am. Are you too? All the best if you are!

@@Rfjcmrvvghbyhb123 Yepp, I have A5 on Monday and A1 on Tuesday. Good luck to you too! Do you feel prepared?

This question for a circuit book and not from PEO exams.

I will do my best to do that.

@@electrical4everyone Hi professor, for Q6 of May 2023 part b, shall I use the equation E= 1/2 C V square? Thank you!

Only at the steady state solution when the switch stays at this position for long time. We are solving a transient problem, so we are interested to find the answer at the switching time and little bit after that.

@@electrical4everyone When Switch is in position A,I ’m confused because there is no source in the circuit. How can we assume that the inductor acts as a short circuit and the capacitor as an open circuit? To determine the initial condition values, we consider the time before 𝑡<0 when the switch was in position A. In this scenario, we analyze the circuit in its steady state to find the initial conditions. Could you explain more please?

@@sureshnagenthirarajah3097 the short circuit model in inductor it means it is fully charged or discarged and the same applies for the open circuit model for the capacitor.

thanks for your nice words

The inductor when full charged will be a short circuit and hence the voltage across it will be zero.

@@electrical4everyoneProfessor since the source is a current. Did you convert it into voltage source with the resistor and then find Vl?

​The source is indeed a voltage source and not a current source.

@@electrical4everyone I think what's being asked is the inductor at t(0+) is being considered a 5A current source. So how did you subtract the 5A current from the 20V voltage source?

​​@@tylerbradburn240I think 5 here is 5v, voltage drop across 1hm resistor, so voltage across the inductor is 20-(5×1)=15V

Thanks for your nice words.

You are very welcome

you are correct thanks for your feedback.

Yes because the 75 ohm is short circuited and hence is cancelled.

@@electrical4everyone Req is calculated without 75 Ohm because it is short-circuited. so we just calculate the total power loss using I^2*Req. Why did you calculate the power loss by each resistor? It is a long process...

I think you need to find it for each resistor to check the power balance. Req =V/I which are calculated to find the power from the source =V*I. hence just finding Req make you repeat the same formula.

@@electrical4everyone Please solve QN. 4 , December 2021. Thanks.

Sure, but it will be much more complicated.

Si le metes Laplace lo vas a complicar mas.

you are absolutely correct

I will try my best

I will try to solve as much as I can

you are welcome

@@electrical4everyone Please upload more videos.

@khagendrathapa8008 This is the plan. I will try to solve as much as I can.

you are welcome

you are welcome

you are welcome

This is the plan. I will start with all different questions.

How can we get PEO PEng questions for 2020, 2021, 2022, and 2023?

@khagendrathapa8008 please email me and I can send all exams I have

@@electrical4everyone Sent you an email. Thanks.

​@electrical4everyone, sir, your email Id, please?

the power cannot be zero. I cannot understand your answer from your description.

@@electrical4everyone I further reduce the circuit, and it seems I can reduce it into 2 nos. 12 V source in the opposite direction.

there is no current in this branch because the capacitor is open circuit.

@electrical4everyone I see. Thank you

@@electrical4everyone it was possible to apply voltage division ?

@assiatinkicht6755 voltage division will not be easy as you need to have all elements in series with the voltage source

you are welcome

you are right, I wrote correction in the video description

Pin can be either = Pout + Pcore + Pcu or Pin = Vp*Ip*cos(theta), both should give you the same answer.

The question is asking for Vo(t) and hence we should solve this problem using peak values and not RMS values. Also this is why I multiplied the real and reactive power formula with (1/2)