I wish I found your channel when I was younger. I am happy I found it now. My major is Mechanical Engineering but read about QM and the history of physics. I made a choice in continuing phyics courses beyond my field of study. I remember seeing you in Brian Greene’s documentary when I was child. I am now 25 years old. Thank you for your generosity in sharing the understanding and helping me uncover the laws of the universe. ❤
(a) Immediately after the switch is closed: The inductor will oppose the current through R₃ and make it initially to zero. => I₃ = 0 and I₁ = I₂ = ℰ/(R₁ + R₂) (b)The switch is closed for a long time: No back EMF left in the inductor. => L acts as a short. Let Rp be the parallell combination of R₂ and R₃ => Rp = R₂·R₃/(R₂ + R₃) Define V’ as the voltage at the point where R₁, R₂ and R₃ meets, with respect to ”ground” or the minus side of the voltage supply. => V’ = ℰ·Rp/(R₁ + Rp) => I₂ = V’/ R₂ and I₃ = V’/R₃ I₁ = I₂ + I₃ (c) Immediately after the switch is open: The inductor will maintain the current through R₃ and drives it through R₂ in the opposite direction than before.
Since the voltage supply is disconnected => I₁ = 0 Current is maintained => I₃ = V’/R₃ = -I₂ (d) The switch has been open for a long time: All energy in the inductor is lost, so all current goes to zero: => I₁ = I₂ = I₃ =0
Bravo! I like this approach - more like the way I would approach it as an engineer, step by step and using intermediate results, than blindly writing every result as a long equation in terms of the R values.
I am curious to learn about your favorite book within the realm of physics or any related subjects. Additionally, I would greatly value any book recommendations you may have that could enrich my understanding of the field Warm regards,
Sir i hope you are well. Please keep your self from this disgusting clever world. One's innocence should never be destroyed. But you already know that :)
You are great, thank you for your good teaching, sorry, I have a question, if we look at the experiment through a mirror in the experiment of throwing electrons, how will the reaction of the electrons be??
Open circuit @ t=0 and short circuit @ t= long time respectively. Using the formula i= v/r(e^t/T). Professor please reply me.....I will be very happy 🙏🙏🙏
Sir can you please upload syllabus of physics(Like chapter by chapter index with subtopics ) generally taught to highschool students in your country😄😄 just curious to known 😅 IN INDIA there is neither education nor system in our education system😓
I was hoping for a question like what is each current 1 second after closing the switch. Then, if you don't know how to apply thevenin theorem, your in a lot of trouble. If you do, of course, still easy. Assuming there was no current in the inductor before closing the switch, I think as follows : a) At t=0 and the switch is closed, the inductor will resist all current change. As it was 0 it will still be 0, so no inductor current and no current I3. Therefore : I1=I2=V/(R1+R2), I3=0. b) After a long time, the inductor will have an impedance of 0 and behave as a perfect conductor. The replacement resistor for R2 and R3 in parallel will be Rr = 1/(1/R2+1/R3). Then I1=V/(R1+Rr). The remaining voltage over R1 and R2 will be : V23 = V-I1*R1. Then I2 = V23/R2 and I3 = V23/R3. c) Starting from answer b, if the switch is opened, the inductor will again resist all current change. I3 will remain at the value specified in answer b. As the switch is open now, current I1 will be 0. What remains is the loop R3, inductor and R2. As I3 remains at it's value, this current will pass through R3 as was the case in answer b and this same current will pass through R2. However, against the arrow drawn in the diagram. So, I2 will be negative. So : I1=0, I3 = I3(answer b) and I2 = -I3. d) After a long time all energy in the system will be converted to heat in the resistors R2 and R3. No current anywhere in the system and no voltage. So I1=I2=I3=0.
That does make an interesting take on this question. If it were my choice, I'd use the Laplace transform to solve it. Inductor impedance Zl = s*L, and resistor impedances simply equal resistances. The switch movements and voltage source are step functions, which in this case would be V(s) = E/s for the first part, and V(s) = E/s - E*e^(-a*s)/s, after the switch reopens at time t=a.
@@lecturesbywalterlewin.they9259 thank u very much sir Sir is there any way we can get these questions in pdf format as it will be easy for us to attempt them
Solutions: A: I1_0 = E/(R1 + R2) I2_0 = E/(R1 + R2) I3_0 = 0 B: I1_ss = E*(R2 + R3)/(R1*R2 + R1*R3 + R2*R3) I2_ss = E*R3/(R1*R2 + R1*R3 + R2*R3) I3_ss = E*R2/(R1*R2 + R1*R3 + R2*R3) C: I1_jo = 0 I2_jo = -E*R2/(R1*R2 + R1*R3 + R2*R3); negative means it is opposite what the picture shows I3_jo = E*R2/(R1*R2 + R1*R3 + R2*R3) D: I1_inf = 0 I2_inf = 0 I3_inf = 0 Supporting Calcs: Assign subscripts: _0 = initially after switch closes _ss = steady state, a long time after switch closes _jo = Just after switch opens _inf = settling state, after infinite time By inspection, I3_0 = 0. Initial inductor current can't abruptly change, from being zero prior to the switch closing. I3_jo = I3_ss. Inductor current is continuous before and after the switch opens. I1_jo = 0. The switch blocks current through the voltage source and R1. I2_jo = -I3_jo. Inductor current has nowhere else to go, once switch opens, so it flows back thru resistor R2. I1_inf = I2_inf = I3_inf = 0. All current eventually dissipates to zero. Initially, with no inductor current, the circuit reduces to R1 and R2 in series. Total resistance is R1 + R2. Divide voltage by total resistance to get current. I1_0 = I2_0 = E/(R1 + R2) At steady state, inductor current in DC is constant. The inductor behaves as if it were a direct connection. The circuit reduces to R2 & R3 in parallel, which is in series with R1. Combine resistances accordingly. R23 = R2*R3/(R2 + R3) Rnet = R1 + R23 Rnet = R1 + R2*R3/(R2 + R3) Construct Ohm's law for the full current, I1: E = I1_ss*Rnet Construct KCL & KVL to find currents in R2 & R3 I2_ss + I3_ss = I1_ss I2_ss*R2 = I3_ss*R3 Solutions to the this system of 3 equations, are at the top. So, thus far, we've found all initial currents, and all steady state currents. By our earlier inspections, all remaining currents are either 0, or determined directly from steady state currents.
a) I1 = I2 = U / (R1 + R2), I3 = 0 b) I1 = U / (R1 + R2*R3/(R2+R3)), I2 = (U - R1*I1)/R2, I3 = (U - R1*I1)/R3 c) I1 = 0, I2 = I3 = (U - R1*I)/R3 with I := U / (R1 + R2*R3/(R2+R3)) (I = same current like I1 in part b or I3 is the same current as I3 in part b) d) I1 = 0, I2 = 0, I3 = 0
I3=0, I1=I2=e/R1+R2 just after closing switch After long time, I1=e(R2+R3)/R1R2+R1R3+R2R3, I3=eR2/(R1R2+R1R3+R2R3) and I2=eR3/(R1R2+R1R3+R2R3) After again opening of switch, I1=0, I2=-I3=eR2/(R1R2+R1R3+R2R3) After long time, I1=I2=I3=0
a) i1= i2 = E/(R1+R2 ) and i3=0 b) first of all net resistance = R1+ R2R3/(R2+R3) So net current = E/ {{R1+ R2R3/(R2+R3)}} so i1 = net current mentioned above i2= E R2 R3 / [ R1+ R2R3/(R2+R3)] [R2+R3] [ R2] i3= E R2 R3 / [ R1+ R2R3/(R2+R3)] [R2+R3] [ R3] c) just after it is opened = the inductor will support the decaying current and hence it must be like in i1 = 0 i2 = 0 But i3 would remain the same as mentioned in b) part d) all the current will be zero
I wish I found your channel when I was younger. I am happy I found it now. My major is Mechanical Engineering but read about QM and the history of physics. I made a choice in continuing phyics courses beyond my field of study. I remember seeing you in Brian Greene’s documentary when I was child. I am now 25 years old. Thank you for your generosity in sharing the understanding and helping me uncover the laws of the universe. ❤
"If you can't do it ,we'll still be friends " was 🥶 COLD
Hi grandpa 👴 I’m here when I saw your lecture on physics. You are truly a great lecturer 🧑🏫 ❤ .
How many grandpas do you have😂😂
Sir i am preparing for JEE
You are my inspiration ❤❤
Easier than CBSE Physics 12th Board 2024
Right bro easier than set 3
😂😂
easier than any set@@ryhno_aakarsh12
(a) Immediately after the switch is closed:
The inductor will oppose the current through
R₃ and make it initially to zero.
=> I₃ = 0 and I₁ = I₂ = ℰ/(R₁ + R₂)
(b)The switch is closed for a long time:
No back EMF left in the inductor.
=> L acts as a short.
Let Rp be the parallell combination of
R₂ and R₃
=> Rp = R₂·R₃/(R₂ + R₃)
Define V’ as the voltage at the point
where R₁, R₂ and R₃ meets, with respect
to ”ground” or the minus side of the
voltage supply.
=> V’ = ℰ·Rp/(R₁ + Rp)
=> I₂ = V’/ R₂ and I₃ = V’/R₃
I₁ = I₂ + I₃
(c) Immediately after the switch is open:
The inductor will maintain
the current through R₃ and drives it
through R₂ in the opposite direction
than before.
Since the voltage supply is disconnected
=> I₁ = 0
Current is maintained
=> I₃ = V’/R₃ = -I₂
(d) The switch has been open for a long time:
All energy in the inductor is lost, so all current
goes to zero:
=> I₁ = I₂ = I₃ =0
Bravo! I like this approach - more like the way I would approach it as an engineer, step by step and using intermediate results, than blindly writing every result as a long equation in terms of the R values.
@@oldtvnut Thank you very much!
Sir May your name be alive for 1000s of years. What a Man u r❤
Walter Lewis is a Paragon when it comes to teaching
A lot of love from india sir
.❤ Thank you so much for creating the love for physics in me. Youll always be the best
Sir you are a living legend of physics
I am curious to learn about your favorite book within the realm of physics or any related subjects. Additionally, I would greatly value any book recommendations you may have that could enrich my understanding of the field
Warm regards,
Lucky to be your friend...
God bless you sir ! You are a wonderful human being :)
You are so kind
Thank you sir. I had questioned you about Alternating current problem. And here is the solution
Your student from nepal❤
Ma ni nepali ho
HAPPY PI DAY, SIR!
Sir i hope you are well. Please keep your self from this disgusting clever world.
One's innocence should never be destroyed. But you already know that :)
You are great, thank you for your good teaching, sorry, I have a question, if we look at the experiment through a mirror in the experiment of throwing electrons, how will the reaction of the electrons be??
you will see the mirror image
cpt?u doing parity test?
Open circuit @ t=0 and short circuit @ t= long time respectively. Using the formula i= v/r(e^t/T).
Professor please reply me.....I will be very happy 🙏🙏🙏
Sir, what do you think about string theory and parallel universes? (I saw you in a documentary today and I was happy as if I saw a familiar person ❤.)
use google
@@lecturesbywalterlewin.they9259why not books? :)
Sir,pls make a video on mirage and why it is not formed due to tir?
@@aritrasaha8-bRead Ncert physics book
Hii sir namaste 🙏 respect from INDIA
Sir can you please upload syllabus of physics(Like chapter by chapter index with subtopics ) generally taught to highschool students in your country😄😄
just curious to known 😅
IN INDIA there is neither education nor system in our education system😓
there are 3 playlists on my channels (with subjects mentioned) 8.01, 8.02 and 8.03
Your we still be friends hits 2 times harder than my girlfriend saying it
I was hoping for a question like what is each current 1 second after closing the switch. Then, if you don't know how to apply thevenin theorem, your in a lot of trouble. If you do, of course, still easy.
Assuming there was no current in the inductor before closing the switch, I think as follows :
a) At t=0 and the switch is closed, the inductor will resist all current change. As it was 0 it will still be 0, so no inductor current and no current I3. Therefore : I1=I2=V/(R1+R2), I3=0.
b) After a long time, the inductor will have an impedance of 0 and behave as a perfect conductor. The replacement resistor for R2 and R3 in parallel will be Rr = 1/(1/R2+1/R3). Then I1=V/(R1+Rr). The remaining voltage over R1 and R2 will be : V23 = V-I1*R1. Then I2 = V23/R2 and I3 = V23/R3.
c) Starting from answer b, if the switch is opened, the inductor will again resist all current change. I3 will remain at the value specified in answer b. As the switch is open now, current I1 will be 0. What remains is the loop R3, inductor and R2. As I3 remains at it's value, this current will pass through R3 as was the case in answer b and this same current will pass through R2. However, against the arrow drawn in the diagram. So, I2 will be negative. So : I1=0, I3 = I3(answer b) and I2 = -I3.
d) After a long time all energy in the system will be converted to heat in the resistors R2 and R3. No current anywhere in the system and no voltage. So I1=I2=I3=0.
That does make an interesting take on this question. If it were my choice, I'd use the Laplace transform to solve it. Inductor impedance Zl = s*L, and resistor impedances simply equal resistances. The switch movements and voltage source are step functions, which in this case would be V(s) = E/s for the first part, and V(s) = E/s - E*e^(-a*s)/s, after the switch reopens at time t=a.
Sir I'm in 8th standard....and I want to know the Schrödinger's equation.........H ^ ψ = E ψ...please explain me about it......
en.wikipedia.org/wiki/Schr%C3%B6dinger_equation
Thank you sir
Best book for rotational motion sir??
For problems: IE Irodov
For theory : University physics
I used them for rotational motion
Sir please tell where can we get other problems like this and the solution to them
195 problems in one playlist 194 solutions so far in another playlist. About 140 problems in the Help Sessions of 8.01 and 8.02
@@lecturesbywalterlewin.they9259 thank u very much sir
Sir is there any way we can get these questions in pdf format as it will be easy for us to attempt them
Sir is there any way we can get these questions in pdf format as it will be easy for us to solve them
@@sanyamgoel4175 I do not know how to post pdf files - In any case I by far prefer videos -
@@lecturesbywalterlewin.they9259 ok sir thank u very much
Hey professor, Once again I have uploaded my attempt to the question.
Please check it out.
Professor you remember what is tomorrow?
Solutions:
A:
I1_0 = E/(R1 + R2)
I2_0 = E/(R1 + R2)
I3_0 = 0
B:
I1_ss = E*(R2 + R3)/(R1*R2 + R1*R3 + R2*R3)
I2_ss = E*R3/(R1*R2 + R1*R3 + R2*R3)
I3_ss = E*R2/(R1*R2 + R1*R3 + R2*R3)
C:
I1_jo = 0
I2_jo = -E*R2/(R1*R2 + R1*R3 + R2*R3); negative means it is opposite what the picture shows
I3_jo = E*R2/(R1*R2 + R1*R3 + R2*R3)
D:
I1_inf = 0
I2_inf = 0
I3_inf = 0
Supporting Calcs:
Assign subscripts:
_0 = initially after switch closes
_ss = steady state, a long time after switch closes
_jo = Just after switch opens
_inf = settling state, after infinite time
By inspection,
I3_0 = 0. Initial inductor current can't abruptly change, from being zero prior to the switch closing.
I3_jo = I3_ss. Inductor current is continuous before and after the switch opens.
I1_jo = 0. The switch blocks current through the voltage source and R1.
I2_jo = -I3_jo. Inductor current has nowhere else to go, once switch opens, so it flows back thru resistor R2.
I1_inf = I2_inf = I3_inf = 0. All current eventually dissipates to zero.
Initially, with no inductor current, the circuit reduces to R1 and R2 in series. Total resistance is R1 + R2. Divide voltage by total resistance to get current.
I1_0 = I2_0 = E/(R1 + R2)
At steady state, inductor current in DC is constant. The inductor behaves as if it were a direct connection. The circuit reduces to R2 & R3 in parallel, which is in series with R1. Combine resistances accordingly.
R23 = R2*R3/(R2 + R3)
Rnet = R1 + R23
Rnet = R1 + R2*R3/(R2 + R3)
Construct Ohm's law for the full current, I1:
E = I1_ss*Rnet
Construct KCL & KVL to find currents in R2 & R3
I2_ss + I3_ss = I1_ss
I2_ss*R2 = I3_ss*R3
Solutions to the this system of 3 equations, are at the top.
So, thus far, we've found all initial currents, and all steady state currents. By our earlier inspections, all remaining currents are either 0, or determined directly from steady state currents.
a) I1 = I2 = U / (R1 + R2), I3 = 0
b) I1 = U / (R1 + R2*R3/(R2+R3)), I2 = (U - R1*I1)/R2, I3 = (U - R1*I1)/R3
c) I1 = 0, I2 = I3 = (U - R1*I)/R3 with I := U / (R1 + R2*R3/(R2+R3)) (I = same current like I1 in part b or I3 is the same current as I3 in part b)
d) I1 = 0, I2 = 0, I3 = 0
I3=0, I1=I2=e/R1+R2 just after closing switch
After long time, I1=e(R2+R3)/R1R2+R1R3+R2R3, I3=eR2/(R1R2+R1R3+R2R3) and I2=eR3/(R1R2+R1R3+R2R3)
After again opening of switch, I1=0, I2=-I3=eR2/(R1R2+R1R3+R2R3)
After long time, I1=I2=I3=0
Am in class 8 so I dont know this stuff
a) I1 = I2 = E/(R1+R2); I3 = 0;
b) I1 = I2 + I3 = (R2+R3)E/(R1(R2+R3)+R2R3); I2 = (E-I1R1)/R2; I3 = (E-I1R1)/R3;
c) I1 = 0; I3 = -I2 = (E1-I1R1)/R3;
d) I1 = I2 = I3 = 0;
"If you can't solve this problem, then..."
prof that was personal I'm just a mere highschooler 🤣😭
i mean so am i
anyone must do that
u dont need to get into college for this
(a) I1=ε/(R1+R2); I2=(ε+R1(ε/R3-R1))/R2; I3=(1/R3)-((ε/R3)(1+R1/R3-R1))
(b) I1=(ε^2)/(R2-εR1); I2=(1/R2)(ε+(R1ε^2)/(R2-εR1)); I3=(1/R3)(ε+(R1ε^2)/(R2-εR1))
(c) I1=0; I2=R3/R2; I3=1
(d) I1=0; I2=infinite; I3= infinite
Don't think the last two answers are correct
a) i1= i2 = E/(R1+R2 ) and i3=0
b) first of all net resistance = R1+ R2R3/(R2+R3)
So net current = E/ {{R1+ R2R3/(R2+R3)}}
so i1 = net current mentioned above
i2= E R2 R3 / [ R1+ R2R3/(R2+R3)] [R2+R3] [ R2]
i3= E R2 R3 / [ R1+ R2R3/(R2+R3)] [R2+R3] [ R3]
c) just after it is opened = the inductor will support the decaying current and hence it must be like
in i1 = 0
i2 = 0
But i3 would remain the same as mentioned in b) part
d) all the current will be zero
this ques is really very easy sir , it's of JEE Mains level
hello hello hello ❤