Practice Problem 8.2 Fundamental of Electric Circuits (Sadiku) 5th Ed

Practice Problem 8.2 Fundamental of Electric Circuits (Sadiku) 5th Ed - Second Order Circuits

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@yousif13ahmed 4 ай бұрын

In minute 4 when you tried to calculate Vr(0+) . I think that Vr(0+) and Vr(0-) are not equal because Vr(0+) is calculated after we put a current source so I don't think the voltage is zero . I know the voltage can't change instantly but that only applies to the capacitor and not the resistor. Please correct my statement if I'm wrong

@andreaspap1816 4 ай бұрын

thats what i believe too. also the way he calculates dVr/dt it seems wrong to me

@ArdiSatriawan 4 ай бұрын

Yup, see other comments

@adityarawat5659 9 ай бұрын

For people worried about the KCL he did at point A( where Ir & Ic are) , none of the current goes to resistor becase capacitor branch is short circuit i.e Vc(0-)=Vc(0+)=0v

@ArdiSatriawan 9 ай бұрын

@JeremyTan-f3u 9 ай бұрын

No capacitor is open circuit. Ic is zero

@sohamagarwal5809 Ай бұрын

11:20 Why is the current in the capacitor at t = 0+ 4 Amps? I mean, it was short-circuited before t = 0

@lucianonogueira5794 3 жыл бұрын

My mistake calculating VR. At node B, iR = iL + 6 (not 4!!). iR =0 and then VR (0+) = 0.

@ArdiSatriawan 3 жыл бұрын

So the values were correct?

@lucianonogueira5794 3 жыл бұрын

@@ArdiSatriawan , Yes and no! the values are correct, but the way you got them I supose is not correct. I was a coincidence. My thought, step by step, was: iL(0-)=-6A, then iL(0+)=-6A vR(0-)=0 I´ll check vR(0+) later vC(0-) 0, then vC(0+)=0 In t=0+ we have 2 corrent sources KCL in node B: iR(0+) = iL(0+) + 6. And so, iR(0+) = -6 + 6 = 0. And finally, vR(0+) = R . iR(0+) = 0 vR(0+) is zero because de corrent iR(0+) is zero. Not because vR(0-) must be the same value on 0+. In my opinion just capacitor must have v(0-)=v(0+). If we had a different value for the right corrent source (10A, for example), iR(0+) = 4 A and vR(0+) must be 20V (and vR(0-) = 0). It´s my opinion....

@santhoshchandchelli6581 Жыл бұрын

@Ardi Satriawan , While we are calculating the diL(0+)/dt we are saying that current through capacitor is zero at t=0+ (because it acts as an open circuit and thats why all the cuurent goes through 5ohm resistor). On the other side while we are calculating the dVc(0+)/dt we are saying that current through capacitor is 4A ( because of Vr(0-)=Vr(0+)=0) at same situation t=0+. how is that possible? please clear my doubt.

@ArdiSatriawan Жыл бұрын

It depends on the previous condition of the capacitor

@AgustinRiquelme441 Жыл бұрын

In problem b), first you said ic=0 and then suddenly you said ic=4A , why?

@ArdiSatriawan Жыл бұрын

Different times

@user-yt4gh7pr1q 3 ай бұрын

4:20에서 전류가 VL쪽으로만 가서 VR과 전압이 같다고 했는데 자세히 설명해주실 수 있나요? VL 0이니까 VR도 0이라는건가 .... VR(0-)와 VR(0+) 가 같은 이유는 뭔가요 저항은 연속성의 원리를 가지고 있지 않는데

@ArdiSatriawan 3 ай бұрын

See the other pinned comments.

@Ryuseok767 2 ай бұрын

저가 생각하기엔 케페시터가 전압은 급격히 안변하지만 전류는 즉각적으로 변하잖아요 그런데 0+일때 유닛함수가 4A가 되는데 딱 바로 열렷을때 직후는 케페시터가 충전되야하니 4A가 케페시터로 흘러서 그런거 아닐까요?? 저도 이제 공부하는 학부생이라 위에 적어놓은건 팩트보단 제 생각이에요

@l.rangga8497 10 ай бұрын

Makasih bngt pak😊😊.boleh saya meminta rekomendasi bapak buku yang bagus untuk teknik elektro belajar mandiri

@ArdiSatriawan 10 ай бұрын

Pake Sadiku, bagus.

@lucianonogueira5794 3 жыл бұрын

Hello... At 9:30 you say voltage in capacitor is 20V. But it´s on t=0+ Vc must be equal Vc(0-), i.e., 0V. So, I think it´s not correct. At 4:55 you say voltage in resistor can not change instantly. I think it´s true just for capacitor. Then, I think VR(0-)=0, and VR(0+)=-10V (KLC at node B --> iR = iL + 4 --> iR = -2A and then VR(0+)=5xiR = -10V). What do you think about it??

@ArdiSatriawan 3 жыл бұрын

My thought was: The voltage of the resistor depends on Vc, which can't change instantly. But let me think about it. 🤔