Source Free RL Circuit || Example 7.4 || Example 7.5 || Practice Problem 7.5 || LCA 7.3(2)(E)
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- Опубликовано: 6 окт 2024
- LCA 7.3(2) (Alexander) - Example 7.4 , Example 7.5 , Practice Problem 7.5
Example 7.4 : The switch in the circuit of Fig. 7.16 has been closed for a long time.
At the switch is opened. Calculate for
Example 7.5 : In the circuit shown in Fig. 7.19, find and i for all time, assuming that the switch was op
Practice Problem 7.5: Determine I, i0 and v0 for all t in the circuit shown in Fig. 7.22.....
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This video explains source free RL circuit in a very simplified manner.
Example 7.4, 7.5, Practice Problem 7.5
(PP 7.5 has some error. Please see • Source Free RL Circuit... )
#ElectricalEngineeringAcademy
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thank you so much sir .. this is the first time i have understood RL circuit pcompletely :)
Very helpful for me 💕💕
Thanks Sir 👍
Valuable lesson
Great Professor. Thank you for your valuable contribution...
Beautiful work
Lots of respect
U may use kvl for finding the current
U may, but it better to use kcl where possible.
Why did you used voltage source (10v)instead of current source when you are doing current divider law.
Please write video time, where you are .
Greatly appreciated sir❤️
Welcome dear. Best wishes.
Sir, in the last question...why is Io and Vo (4/3) and (8/3).
Where did the 1/3 come from?
My fault, I should have explained it in more details. At time t=14:05. Current i(t) is 4e^-2t (I hope this is clear). Now using CDR, i0(t) = -( i(t)/(6+3))* 3(opposite arm) = -i(t)/3 = - 4/3e^-2t (minus sign is used because current direction of i0(t) is opposite of i(t). v0(t) =-i0(t)*2.
ruclips.net/video/ps6ZQfBY7lA/видео.html
Find the inductor current in the circuit 2 for all time
an
5H
10 A
20
30
Doo this sir plz
Shouldn't v(o) in the last question be positive? Love ur vids btw, helped me so much in understanding electronics 😁
Yes, U R right. Thanks dear.
Beautiful
Excellent
Helpful
Sir, please explain vo(t)= 4e^-t & io(t)= -2/3e^-t details...
Please watch this ruclips.net/video/GUVWh82dIu4/видео.html
Good
Wonderful
Good job
SIR CAN YOU EXPLAIN WHY V nought = negative vl?
at time 09:10, I have explained it. Just redraw the diagram with 3ohm resistor in parallel to 6ohm and Inductor .
Thank you sir make more videos
Can you explain how 4 ohm and 12 ohm resistors are in parallel bcoz apperantly they look like in series please reply as soon as possible bcoz I need the solution for my exams
Where did I say that ?
@@ElectricalEngineeringAcademy in the start of video u solved 4 and 6 ohm resistors as a parallel resistor but they are in series and in our book they also solved it in parallel even they are in series
@@Bluemoon11-7v watch this to clear you concept
ruclips.net/video/xGmqp2JH_NI/видео.html
3:44 and 8:31 I am very confused between example 1 and 2. For example 1, when t
Listen to the video very carefully from 0801 time. Your confusion is as to why we are not shorting the whole right hand side. Actually the circuit should be looked into as two circuits : one is, the 10V, and 2 ohm, and the short circuit, the other is the short circuit and whole of right hand side. In the first case we will get a current 10/2=5A, but this current is not through the inductor, hence of no use to us. So we now look at the circuit on the right hand side. The circuit becomes as what is shown at 8:37. Note the short is not directly across 6 ohm, hence 6 ohm can not get eliminated.
Also can u please explain how Vo(t)=-8/3*e^-2t
Why 16ohm resistor is short circuited
In case of dc (steady state condition) inductor becomes short circuited. Since the inductor becomes short circuited , and 16 ohm resister will become useless since all current will flow through the short circuited branch. That is why loosely we call that the 16 ohm resistor has become short circuited (actually it is by passed, or become virtually non existent). Another way of looking at it is that short circuit has zero resistance, So R parallel Zero, will give Requivalent = 0.
Beautiful😍
Nice
which book is he using ??
Fundamentals of Electric Circuits by Alexander & Sadiku
@@ElectricalEngineeringAcademy thanks
Sir in first sum u have taken at to u have taken switch is closed why?
Dear Susarla, in future if u give reference of the video time (where u noted a problem), it will be easier for me to find out. Regarding you question, please look carefully the switch arrow direction in both the problems. In 1st sum it opens at t=0, that means it is closed for t
Tqq sir...and sure from next tym I'll follow
last question why i0(t) and v0(t) have - ?
Current io is obtained from current i using kcl equation. If u look at right most diagram, u will notice that the direction of current io is opposite of current i, hence the negative sign.
watch this ruclips.net/video/ps6ZQfBY7lA/видео.html
if we have to find i(t) for t=10ms what we have to do ?
Put t=0.01 in final equation and solve
Keep uploading more videos
wow!!!!
2:51 why the 16ohm resistor short-circuited?
When DC current flows through an inductor, it behaves like a short circuit. (except for transient period or very very short time).
@@ElectricalEngineeringAcademy yes i know that, some people forget about the node, its because the node that makes the resistor is short-circuited because the current flows through the inductor is same as the resistor 16ohm. i just got the explanation.
@@ElectricalEngineeringAcademybut why on the 2nd question, the circuit is open but the inductor as short cuircuit?
@@vaniandriyani Please write video time, so I can follow you.
@@vaniandriyani please write video time, so I can follow you
sir kindly,
tell how to i(o) and v(o) of self test question no.3
Please give reference to the time scale of the video where you are having trouble.
Gracious
Thanks Khalifa. Best Wishes
Sir how did we get I(1)=12/12+4*I(0)
By using current division rule CDR, according to which current i1(0) (in branch having resistance R=4) is equal to = (total current i(0))divided by total resistance of the two parallel branches (12+4), and multiplied by the opposite arm resistance 4ohm. So i1(0)= (i(0)/(12+4))*4 , which can also be written as (4/(12+4))*i(0). If u can't follow just see my circuit1 video for CDR (current division rule).
At 14:06 how did 8 become 8/3? should it be 8e^-2t?
I guess there is a mix-up. In RL circuit, to calculate i(t), we use i(0) value i.e i(t) = i(0)e^-t/tau, but v(t) is not calculated as v(0)e^-t/tau; it is rather calculated as v(t) = R x i(t) = 2 x (-4/3) e^-2t. In RC circuit the opposite happens; that is v(t) is calculated from v(t)=v(0) e^-t/tau (Just for info, there is however an error in the sign of v(t) in the answer; it should be (-)x(-) = +.)
ruclips.net/video/TEIzOBMoz1s/видео.html
wow
Sir last question nhi samajh aaya
ruclips.net/video/ua5vKecgto4/видео.html
ruclips.net/video/pwwAqZ_Vu7Y/видео.html
Garat
nc
for problem 7.4
Do you mean Practice Problem 7.4 ?
ruclips.net/video/sujdIDI7O9A/видео.html
Hindi video nahi ha sir
Aik aad din me upload kardoonga, Inshaa Allah
Aap ke liye bahot sare urdu/hindi videos load kar dia hai. Aaapne like nahi kia ? apne dosto me bhi share karen.
Worst
Thanks dear; I tried.
Beautiful
Excellent
Nice
Good
Also can u please explain how Vo(t)=-8/3*e^-2t
vo(t) is the voltage across 2 ohm resistor. v0(t) = i0(t) x 2 (ohm)
Yes sir...but to find io(t) we have to find vo(t) that is vo(t)=VL=L*di/dt...and where i=i(t)=4*e^-2t which we obtained firther so from the above vo(t) should be -8*e^-2t
Oh dear, I should have explained it , but thought that after having done the earlier questions, one would understand it. Any way , look at the last diagram, we got i(t) by using formula i(t)=I(0)e^(-t/tau) =4*e^-2t. The i(t) is dividing into two branches: the upper branch and the lower branch. We are interested in the lower branch current.So we used Current Division Rule CDR to find current Io in the lower branch Io= - i(t) divide by total resistance (3+6) and then multiplying by Resistance of opposite arm (3 ohm) that is - 4*e^-2t x 3/(3+6) = -4/3*e^-2t (negative sign because the direction of the current in lower branch due to i(t) is actually opposite of the direction of current Io). Now we use this current to find voltage across 2 ohm resistance by multiplying by 2, so Vo(t)=-8/3*e^-2t
@@ElectricalEngineeringAcademy Padu sir 👍👍👍
@@ElectricalEngineeringAcademy padu sir 👍👍👍
Beautiful