The graph will be a combination of exponential graph(one of which will be enforced at a time) changing sign of its slope at equal time intervals and its amplitude will also keep decreasing and become constant as t approaches infinity To find the area under v-t curve(integration) We use considering clockwise as + direction V=Ldi/dt+Ri Vdt=Ldi+Ridt Now for the right part we need the area under curve of the Vout-t graph and on dividing it by R we’ll get idt which is area under curve of i-t graph and we can find instantaneous current from instantaneous Vout this way we can find the lhs
I must say, it's not very clear to me what exactly the question is. However, I deal with this kind of circuit all the time. First of all, If you apply a voltage to this circuit, as long as there is no change in the voltage, there will be exponential decay up to the limit of the applied input voltage. This will also be the case for this circuit. However, if the RL time constant is much higher than the period, we are dealing with a small piece of this curve and the exponential decay will not show but it will give us a more or less linear slope. So what you see on the output will be a sawtooth signal. Going up with positive input and going down with the negative. However, the output signal will be small compared to the input signal cause we are dealing with the beginning and tiny part of this curve. Note that it will stay above zero ( almost ) as the input voltage begins with a positive voltage. As the negative voltage starts the slope will go negative but since the input signal duty cycle is 50% it will go to zero again. So it will conform to a perfect integrator for the naked eye in this first few periods. As time passes however, if you put the output on an oscilloscope and look at a time scale several times the time constant you will find that the average output voltage will approach the average of the input voltage, simply according to Uout(average) = Uin(average)*(1-e^(-t/Tau)) and the sawtooth signal will be above zero and below zero for the same amount, averaging to zero as the input signal averages to zero. So, that describes what will happen. No difficult math in there however. I must note that this circuit has enormous application in modern electronics. As the world changed from analog to digital electronics, this very circuit converts digital signals to analog again. If you have a computer with sound, this circuit is what generates the sound. Same for a modern amplifier, car radio, electronic organ, you name it. ( although it will be mostly in the form of an RC circuit but same principles apply )
In modern electronics, this circuit has 0, absolutely 0 application. Please don't preach out of your own ignorance. This circuit is useful for students learning the material and was maybe useful back in the early days of electronics, but it is completely archaic. In a modern design, that isn't some little hobbyist project, you will never ever see an LR lowpass filter.
@@fredthechamp3475 I agree, rarely. As I said : it will be mostly in the form of an RC circuit but same principles apply. Raspberry pi use and RC filter to generate audio. Countless amplifiers, countless DSP circuits. That's not my ignorance, thats how it is. But I agree, an LR lowpass filter is rare but for example a class D audio amplifier output stage will have such a circuit, if you count the loudspeaker as R which is fair to say imho.
Trying to piece together coherence from everywhere prof because I know nothing about this really and I dont have time because I need to prepare for my/work tomorrow. Constructive and Destructive Interference is really stuck in my imagination for some reason🦅also after seeing one of the videos to help that there is 1Resistance, 2Impedance and 3Reactance at play in the RLC circuits🧐Anyway😉I think these videos are helpful for anyone trying to solve: 1) RLC Circuit Differential Equation | Lecture 25 | Differential Equations for Engineers 2) Integrator - Operational Amplifier | Basic Circuits #14 3) RLC Circuits - Differential Equation Application 4) Kirchhoff's Current Law (KCL) - How to Solve Complicated Circuits 5) Lesson 9.1 - Second Order Differential Equations ChatGPT's response just for interest sake😅 If a resistor is inserted in a closed circuit(Vin - Inductor - Vout - Vin) between Incuctor(L) and after Vout, how would current be influenced by L and R? ChatGPT: When a resistor (R) is inserted into a closed circuit with an inductor (L) and a voltage source (Vin) before and after Vout, the behavior of the circuit changes due to the addition of the resistor. Here's how the current in the circuit would be influenced by the inductor (L) and the resistor (R): Initial Conditions: Initially, when the circuit is first powered on, the inductor will resist changes in current flow. It will initially behave like a short circuit, allowing current to flow freely. However, as time passes, the inductor will build up a magnetic field and resist changes in current. Inductor Behavior: The inductor will still oppose changes in current flow as before. It will produce a back electromotive force (EMF) that opposes the flow of current. This back EMF is proportional to the rate of change of current (V = L di/dt, where V is the voltage across the inductor, L is the inductance, and di/dt is the rate of change of current). Therefore, the inductor will continue to smooth out changes in current flow and resist sudden changes. Resistor Influence: The resistor, being a passive component, will dissipate power in the form of heat according to Ohm's law (V = IR, where V is the voltage across the resistor, I is the current flowing through it, and R is the resistance). The presence of the resistor will introduce a voltage drop in the circuit proportional to the current flowing through it. Steady State: In the steady state (when the circuit has reached a stable condition), the current flowing through the circuit will be influenced by both the inductor and the resistor. The inductor will still try to maintain a continuous flow of current, while the resistor will limit the current flow by dissipating energy. Time Response: The time response of the circuit, particularly during transient conditions (such as when the circuit is first turned on or when there are sudden changes in input), will be affected by both the inductor's tendency to resist changes in current and the resistor's damping effect on the system. In summary, the insertion of the resistor will affect the behavior of the circuit by introducing additional resistance to the flow of current, thereby influencing the overall current in the circuit along with the inductor's behavior.
Sir ! I am a JEE 2025 as well as NEET 2025 aspirant from india and during my preparation your lectures helped me a lot I completed your 8.01x and 8.02x. All your experiments and teachings have added a value in my life, I feel very confident especially when I am able to complete PYQ of JEE Advanced just by watching your lectures.Now I am able to understand Physics concepts in more depth and Now I do physics "For the love of physics":) Thankyou sir .Wishing your son and great grandchild a happy birthday.
This is the Giancoli 3rd Ed text book supplement solution (Chapter 30, p. 14, prob 57: p. 607): From the circuit we see that: Vin = iR + L di/dt, or Ri/L + di/dt = Vin /L. We multiply both sides by the integrating factor exp(Rt/L): exp(Rt/L)Ri/L + exp(Rt/L) di/dt = d(i exp(Rt/L))/dt = exp(Rt/L)Vin /L. If we assume i = 0 when t = 0, when we integrate we get: ₒ∫⁼ͭ d (i·exp(Rt’/ L))= i·exp(Rt/ L) = (1/L)·ₒ∫⁼ͭexp(Rt’/ L)·Vin dt’. If t « L/R, the exponent is small, exp(Rt/L)~1 and exp(Rt’/L)~1, so we have: i = (1/L)·ₒ∫⁼ͭ Vin dt’ Thus the output voltage is: Vout = iR = (R/L) ₒ∫⁼ͭVin dt’. This is the area under the curve when Vin is plotted against Rt/L: For the square wave input Vin, we get: __ __ __ |_| |_| |_| For the output Vout, we get: ╱ ╲ ╱ ╲ ╱ ╲ ╱
Given the time constant is >> waveform time, Vr can be assumed to be so near zero that applied voltage appears in its entirety across the inductor. Output wave is a triangle starting at zero linearly sloping up until zero crossing of Vin, then sloping down to zero with negative of the rising slope, meeting zero at the next zero crossing of the input voltage. Wave is thereafter periodic. Peak voltage is VtR/L where t is the time of a half-cycle of Vin. In reality, because Vin has no DC component, & the initial conditions demand the output to have a DC component, the output triangle wave will, after a time >> L/R, shift downward so the triangle wave is symmetrically set about zero, alternating to +/- VtR/2L. This type of excitation (or sine wave, started at the zero crossing), when applied to a large transformer, lightly loaded or unloaded, will cause the initial magnetizing flux wave peak to be double the normal, steady-state amount. This will cause the transformer's core to saturate & a large surge current, 10x to 50x typical full-load current, to flow, often tripping protective breakers. To avoid this, the wave should be started at it PEAK, contrary to conventional thinking for "zero-crossing switching".
He said several times that the E-field is zero in the inductor and that any voltage detected is actually from the B-field inducing a voltage in the voltmeter. If E=0, then what is V..? 🤔
1) Explain how this integrator works. The input square wave voltage, Vᵢ goes between -A and A volts. So peak-to-peak value is 2A.
If L/R is large compared with the period of Vᵢ we will see a trangle wave at the output with small amplitude. If L/R is small, the output voltage is no longer linear, but describes an exponental curve up on one half cycle an down the next half cycle. 2) Use Farady’s Law. ∮E⋅dl = 0 + Ri(t) - Vᵢ = -L (di/dt) …(1) 3) Solve the differential equation.
- Rearrange eq. (1) => di/dt + (R/L)i(t) = Vᵢ/L - Find an integrating factor (I.F.): e^(∫(R/L)dt) = e^(Rt/L) (this is the hint given in the problem) - Multiply all terms with this I.F: => e^(Rt/L)di/dt + e^(Rt/L)(R/L) i(t) = e^(Rt/L) Vᵢ/L …(2) => d/dt[i(t)·e^(Rt/L)] = e^(Rt/L) Vᵢ/L …(3) (from eq. (2) to eq. (3) we have used the derivative of a product backwards and we can now integrate both sides). => i(t) · e^(Rt/L) = (1/L) ∫[Vᵢ · e^(Rt/L)] dt from t₁ to t₂. => i(t) · e^(Rt/L) = (1/L) = (Vᵢ/L)·(L/R)·[e^(Rt₂/L)- e^(Rt₁ /L)] => i(t) = (Vᵢ/R)·[e^(Rt₂/L)- e^(Rt₁ /L)] · e^(-Rt/L) V₀ = R·i(t) = Vᵢ·[e^(Rt₂/L)- e^(Rt₁ /L)] · e^(-Rt/L) (Not a perfect answer, but this is what I have so far. I leave it to Keith to explain it better 🙂.)
I'm going to skip doing exactly what you asked, and not invoke Faraday's law, but go directly to the transient response of an LR circuit, Vout= Vin(1-e^(-tR/L)). Next let a cycle length = 2T, and invoke superposition of a constant +Vin starting at T=0, plus a voltage of -2Vin starting at t=T. Vout at the end of one cycle is Vin(1-e^(-2TR/L)) -2Vin(1-e^(-TR/L)). If the value of R/L is small, the ratio of (1-e^(-TR/L))/(1-e^(-2TR/L)) approaches 1/2, so the ending voltage approaches zero, the integral of the input waveform over one cycle. The voltage across the resistor is a close approximation (with a scale factor) of the integral of the input. In fact, if the value of R could be made zero, and the output was taken as the current in the (ideal) inductor, we would have a perfect integration. Side note: in cathode ray tube horizontal scanning circuits, a switched voltage is applied across the (non ideal) horizontal yoke coil to get an (approximately) linearly increasing current vs. time, and thus a linearly increasing magnetic field, to deflect the electron beam from left to right.
MY NOTATION: ========= v = v_in w= v_out --------------- SOLUTION: ========= Applying Faradays's law, NOT KIRCHOFF'S, we get the differential equation v-Li' = iR (derived also in Lecture 20 of 8.02) Dividing both sides by L we get v/L - i' = I R/L and now we leave the v/L term alone in the left hand side v/L = i R/L + i' As the hint of the problem says, we multiply both terms of this differential equation by e^(Rt/L) making the RHS the derivative of a product, i.e. e^(Rt/L) v/L = d/dt (product) where product=e^(Rt/L) times i(t). Integrating both sides and using the the Fundamental Theorem of Calculus on the RHS, we get 1/L integral{ e^(Rt/L) v} = product We can now substitute i(t) for w/R 1/L integral{ e^(Rt/L) v} = (w/R) e^(Rt/L) A few more steps we'll get us to w(t) = integral{ v(t) dt } Thus the label ''integrator''. Thus, if v(t) is the input function given in the problem, the output would be a triangular (positive=above the horizontal axis) function whose period is the same as the input.
You solve JEE questions. Now, try questions of TIFR & JEST exams Sir. We Physicists deal with this papers to get into top Science Research Institutes like TIFR, IISER in India.
Hi sir please reply I completed mechanics from you I am targeting jee and Olympiad Are they sufficient and I did you hw assignment!!! Please sir reply!!!! Wishing happy birthday to your son !
@@lecturesbywalterlewin.they9259 This, I think. _____ | + |.....V=0 line - |___| "All on" + "All off"...But you wouldn't want to hang around at zero if you don't want some "glitch" to occur. Makes me think of Ivor Catt's book "The Glitch"... 🖥️0️⃣.1️⃣.0️⃣•5️⃣...😵🛬🔥😱
![Eminem - Houdini [Official Music Video]](http://i.ytimg.com/vi/22tVWwmTie8/mqdefault.jpg)
Happy birthday to your son! Thank you for the lecture!
Happy Holi professor,
Today, on 25 March 2024, the festival of colors (Holi) is being celebrated in India.
Happy Holi ! Mr. Walter
Thanks prof. For uploading such amazing videos and concepts
The graph will be a combination of exponential graph(one of which will be enforced at a time) changing sign of its slope at equal time intervals and its amplitude will also keep decreasing and become constant as t approaches infinity
To find the area under v-t curve(integration)
We use considering clockwise as + direction
V=Ldi/dt+Ri
Vdt=Ldi+Ridt
Now for the right part we need the area under curve of the Vout-t graph and on dividing it by R we’ll get idt which is area under curve of i-t graph and we can find instantaneous current from instantaneous Vout this way we can find the lhs
I must say, it's not very clear to me what exactly the question is. However, I deal with this kind of circuit all the time. First of all, If you apply a voltage to this circuit, as long as there is no change in the voltage, there will be exponential decay up to the limit of the applied input voltage. This will also be the case for this circuit. However, if the RL time constant is much higher than the period, we are dealing with a small piece of this curve and the exponential decay will not show but it will give us a more or less linear slope. So what you see on the output will be a sawtooth signal. Going up with positive input and going down with the negative. However, the output signal will be small compared to the input signal cause we are dealing with the beginning and tiny part of this curve. Note that it will stay above zero ( almost ) as the input voltage begins with a positive voltage. As the negative voltage starts the slope will go negative but since the input signal duty cycle is 50% it will go to zero again. So it will conform to a perfect integrator for the naked eye in this first few periods. As time passes however, if you put the output on an oscilloscope and look at a time scale several times the time constant you will find that the average output voltage will approach the average of the input voltage, simply according to Uout(average) = Uin(average)*(1-e^(-t/Tau)) and the sawtooth signal will be above zero and below zero for the same amount, averaging to zero as the input signal averages to zero.
So, that describes what will happen. No difficult math in there however.
I must note that this circuit has enormous application in modern electronics. As the world changed from analog to digital electronics, this very circuit converts digital signals to analog again. If you have a computer with sound, this circuit is what generates the sound. Same for a modern amplifier, car radio, electronic organ, you name it. ( although it will be mostly in the form of an RC circuit but same principles apply )
In modern electronics, this circuit has 0, absolutely 0 application. Please don't preach out of your own ignorance. This circuit is useful for students learning the material and was maybe useful back in the early days of electronics, but it is completely archaic. In a modern design, that isn't some little hobbyist project, you will never ever see an LR lowpass filter.
@@fredthechamp3475 I agree, rarely. As I said : it will be mostly in the form of an RC circuit but same principles apply. Raspberry pi use and RC filter to generate audio. Countless amplifiers, countless DSP circuits. That's not my ignorance, thats how it is. But I agree, an LR lowpass filter is rare but for example a class D audio amplifier output stage will have such a circuit, if you count the loudspeaker as R which is fair to say imho.
Happy Birthday to your son and great grandson!! 💖
Respect to you, Walter.
Trying to piece together coherence from everywhere prof because I know nothing about this really and I dont have time because I need to prepare for my/work tomorrow. Constructive and Destructive Interference is really stuck in my imagination for some reason🦅also after seeing one of the videos to help that there is 1Resistance, 2Impedance and 3Reactance at play in the RLC circuits🧐Anyway😉I think these videos are helpful for anyone trying to solve: 1) RLC Circuit Differential Equation | Lecture 25 | Differential Equations for Engineers 2) Integrator - Operational Amplifier | Basic Circuits #14 3) RLC Circuits - Differential Equation Application 4) Kirchhoff's Current Law (KCL) - How to Solve Complicated Circuits 5) Lesson 9.1 - Second Order Differential Equations
ChatGPT's response just for interest sake😅
If a resistor is inserted in a closed circuit(Vin - Inductor - Vout - Vin) between Incuctor(L) and after Vout, how would current be influenced by L and R?
ChatGPT:
When a resistor (R) is inserted into a closed circuit with an inductor (L) and a voltage source (Vin) before and after Vout, the behavior of the circuit changes due to the addition of the resistor.
Here's how the current in the circuit would be influenced by the inductor (L) and the resistor (R):
Initial Conditions: Initially, when the circuit is first powered on, the inductor will resist changes in current flow. It will initially behave like a short circuit, allowing current to flow freely. However, as time passes, the inductor will build up a magnetic field and resist changes in current.
Inductor Behavior: The inductor will still oppose changes in current flow as before. It will produce a back electromotive force (EMF) that opposes the flow of current. This back EMF is proportional to the rate of change of current (V = L di/dt, where V is the voltage across the inductor, L is the inductance, and di/dt is the rate of change of current). Therefore, the inductor will continue to smooth out changes in current flow and resist sudden changes.
Resistor Influence: The resistor, being a passive component, will dissipate power in the form of heat according to Ohm's law (V = IR, where V is the voltage across the resistor, I is the current flowing through it, and R is the resistance). The presence of the resistor will introduce a voltage drop in the circuit proportional to the current flowing through it.
Steady State: In the steady state (when the circuit has reached a stable condition), the current flowing through the circuit will be influenced by both the inductor and the resistor. The inductor will still try to maintain a continuous flow of current, while the resistor will limit the current flow by dissipating energy.
Time Response: The time response of the circuit, particularly during transient conditions (such as when the circuit is first turned on or when there are sudden changes in input), will be affected by both the inductor's tendency to resist changes in current and the resistor's damping effect on the system.
In summary, the insertion of the resistor will affect the behavior of the circuit by introducing additional resistance to the flow of current, thereby influencing the overall current in the circuit along with the inductor's behavior.
Sir ! I am a JEE 2025 as well as NEET 2025 aspirant from india and during my preparation your lectures helped me a lot I completed your 8.01x and 8.02x. All your experiments and teachings have added a value in my life, I feel very confident especially when I am able to complete PYQ of JEE Advanced just by watching your lectures.Now I am able to understand Physics concepts in more depth and Now I do physics "For the love of physics":) Thankyou sir .Wishing your son and great grandchild a happy birthday.
yes 8.02x is really helpful
im preparing for jee2025 ass well
Excellent !
V output is one unit shifted downward compared to Vinput ... Nature of output graph is same as of V input
This is the Giancoli 3rd Ed text book supplement solution
(Chapter 30, p. 14, prob 57: p. 607):
From the circuit we see that:
Vin = iR + L di/dt, or Ri/L + di/dt = Vin /L.
We multiply both sides by the integrating
factor exp(Rt/L):
exp(Rt/L)Ri/L + exp(Rt/L) di/dt
= d(i exp(Rt/L))/dt = exp(Rt/L)Vin /L.
If we assume i = 0 when t = 0, when we integrate we get:
ₒ∫⁼ͭ d (i·exp(Rt’/ L))= i·exp(Rt/ L)
= (1/L)·ₒ∫⁼ͭexp(Rt’/ L)·Vin dt’.
If t « L/R, the exponent is small,
exp(Rt/L)~1 and exp(Rt’/L)~1,
so we have:
i = (1/L)·ₒ∫⁼ͭ Vin dt’
Thus the output voltage is:
Vout = iR = (R/L) ₒ∫⁼ͭVin dt’.
This is the area under the curve when
Vin is plotted against Rt/L:
For the square wave input Vin, we get:
__ __ __
|_| |_| |_|
For the output Vout, we get:
╱ ╲ ╱ ╲ ╱ ╲ ╱
A lot of respect to you sir❤
LOVE YOU ARE WORKS SIR
Given the time constant is >> waveform time, Vr can be assumed to be so near zero that applied voltage appears in its entirety across the inductor. Output wave is a triangle starting at zero linearly sloping up until zero crossing of Vin, then sloping down to zero with negative of the rising slope, meeting zero at the next zero crossing of the input voltage. Wave is thereafter periodic. Peak voltage is VtR/L where t is the time of a half-cycle of Vin.
In reality, because Vin has no DC component, & the initial conditions demand the output to have a DC component, the output triangle wave will, after a time >> L/R, shift downward so the triangle wave is symmetrically set about zero, alternating to +/- VtR/2L.
This type of excitation (or sine wave, started at the zero crossing), when applied to a large transformer, lightly loaded or unloaded, will cause the initial magnetizing flux wave peak to be double the normal, steady-state amount. This will cause the transformer's core to saturate & a large surge current, 10x to 50x typical full-load current, to flow, often tripping protective breakers. To avoid this, the wave should be started at it PEAK, contrary to conventional thinking for "zero-crossing switching".
He said several times that the E-field is zero in the inductor and that any voltage detected is actually from the B-field inducing a voltage in the voltmeter. If E=0, then what is V..? 🤔
this was decently easy, sir please do try the jee adv 2022, problem on chimeny/incompressible flow
Professor,
You have predicted that you will get three correct solutions, but I will make sure that you do not get only three responses.
1) Explain how this integrator works.
The input square wave voltage, Vᵢ
goes between -A and A volts.
So peak-to-peak value is 2A.
If L/R is large compared with the period
of Vᵢ we will see a trangle wave at the
output with small amplitude.
If L/R is small, the output voltage is no
longer linear, but describes an exponental
curve up on one half cycle an down the
next half cycle.
2) Use Farady’s Law.
∮E⋅dl = 0 + Ri(t) - Vᵢ = -L (di/dt) …(1)
3) Solve the differential equation.
- Rearrange eq. (1)
=> di/dt + (R/L)i(t) = Vᵢ/L
- Find an integrating factor (I.F.):
e^(∫(R/L)dt) = e^(Rt/L)
(this is the hint given in the problem)
- Multiply all terms with this I.F:
=> e^(Rt/L)di/dt + e^(Rt/L)(R/L) i(t) = e^(Rt/L) Vᵢ/L …(2)
=> d/dt[i(t)·e^(Rt/L)] = e^(Rt/L) Vᵢ/L …(3)
(from eq. (2) to eq. (3) we have used the derivative
of a product backwards and we can now integrate both sides).
=> i(t) · e^(Rt/L) = (1/L) ∫[Vᵢ · e^(Rt/L)] dt from t₁ to t₂.
=> i(t) · e^(Rt/L) = (1/L) = (Vᵢ/L)·(L/R)·[e^(Rt₂/L)- e^(Rt₁ /L)]
=> i(t) = (Vᵢ/R)·[e^(Rt₂/L)- e^(Rt₁ /L)] · e^(-Rt/L)
V₀ = R·i(t) = Vᵢ·[e^(Rt₂/L)- e^(Rt₁ /L)] · e^(-Rt/L)
(Not a perfect answer, but this is what I have so far. I leave it to Keith to explain it better 🙂.)
Thanks for hearing... My request
Thanks Sir for posting a good question
Ty
Great problem
Respect button for Walter lewin sir.
👇
I'm going to skip doing exactly what you asked, and not invoke Faraday's law, but go directly to the transient response of an LR circuit, Vout= Vin(1-e^(-tR/L)).
Next let a cycle length = 2T, and invoke superposition of a constant +Vin starting at T=0, plus a voltage of -2Vin starting at t=T.
Vout at the end of one cycle is Vin(1-e^(-2TR/L)) -2Vin(1-e^(-TR/L)).
If the value of R/L is small, the ratio of (1-e^(-TR/L))/(1-e^(-2TR/L)) approaches 1/2, so the ending voltage approaches zero, the integral of the input waveform over one cycle. The voltage across the resistor is a close approximation (with a scale factor) of the integral of the input. In fact, if the value of R could be made zero, and the output was taken as the current in the (ideal) inductor, we would have a perfect integration.
Side note: in cathode ray tube horizontal scanning circuits, a switched voltage is applied across the (non ideal) horizontal yoke coil to get an (approximately) linearly increasing current vs. time, and thus a linearly increasing magnetic field, to deflect the electron beam from left to right.
A simple opamp circuit would allow for a resistor value of 0 and still get an output voltage thus forming a near perfect integrator.
MY NOTATION:
=========
v = v_in
w= v_out
---------------
SOLUTION:
=========
Applying Faradays's law, NOT KIRCHOFF'S, we get the differential equation
v-Li' = iR (derived also in Lecture 20 of 8.02)
Dividing both sides by L we get
v/L - i' = I R/L
and now we leave the v/L term alone in the left hand side
v/L = i R/L + i'
As the hint of the problem says, we multiply both terms of this differential equation by e^(Rt/L) making the RHS the derivative of a product, i.e.
e^(Rt/L) v/L = d/dt (product)
where product=e^(Rt/L) times i(t). Integrating both sides and using the the Fundamental Theorem of Calculus on the RHS, we get
1/L integral{ e^(Rt/L) v} = product
We can now substitute i(t) for w/R
1/L integral{ e^(Rt/L) v} = (w/R) e^(Rt/L)
A few more steps we'll get us to
w(t) = integral{ v(t) dt }
Thus the label ''integrator''. Thus, if v(t) is the input function given in the problem, the output would be a triangular (positive=above the horizontal axis) function whose period is the same as the input.
I solved it on my own n got the same, how would we know if its right?
Everything has started to align , it was seen in full moon cloud card .
Tf
Jee advance problem or jee advance level problem?
❤❤❤❤
Professor, I am chronically depressed what should I do? (please don't say use google)
talk to your doctor
🤣🤡😈☠️
@@lecturesbywalterlewin.they9259😂
@@lecturesbywalterlewin.they9259savage
@@lecturesbywalterlewin.they9259😂😂
You solve JEE questions. Now, try questions of TIFR & JEST exams Sir. We Physicists deal with this papers to get into top Science Research Institutes like TIFR, IISER in India.
Respect for waltner lewin❤
👇
Spell his name right first!
question bana?
Where to send the solution?
Hello professor
Welcome🙋♂️
What edition of Giancoli?
WL uses edition 3 I think.I believe other editions contain the problems, but the numbering is different.
Hi sir please reply
I completed mechanics from you I am targeting jee and Olympiad
Are they sufficient and I did you hw assignment!!!
Please sir reply!!!!
Wishing happy birthday to your son !
let me be honest
the world is sufficient for your needs but not your greeds
Tq
What is your age❓
👆 The “Respect WL button” is up there,
on the left...so go hammer 🔨 that button.
Also: 🎂🎉🎈 HB for WL's son.
you were right. there are not 2.7 cycles. There are a bit more than 3 - with my apologies.
@@lecturesbywalterlewin.they9259
This, I think.
_____
| + |.....V=0 line
- |___|
"All on" + "All off"...But you wouldn't want to hang around at zero if you don't want some "glitch" to occur. Makes me think of Ivor Catt's book "The Glitch"...
🖥️0️⃣.1️⃣.0️⃣•5️⃣...😵🛬🔥😱
Sir what is time ???
use google
No, sir you can't understand my question
According to quantum mechanics time is just a illusion .
What do you believe ?
@@ayushmansharma-lp8te Time equals Distance over Velocity.
@@_John_Sean_Walker😂😂
@@lecturesbywalterlewin.they9259😂
Is this actually a previously asked question of JEE A or are you categorising this as one
where can we send solutions to the problem
Post as a new comment.
The answers not one that i can post here i guess it has a graph and some and diffrential equations with integration limits
@@lakshatchoudhary You will need to describe the output shape in words not a graph.
Ok sending
sir, ,how do i buy your lectures in india? im 14 years old and i love ur lectures and happy birthday to your son!
does keith have an answer?
Yes but it is a secret for now.
@@KeithandBridgetIs V=0 at t=0 or is it Vmax ❔😕
@@The_Green_Man_OAP At t=0 it transitions from 0 to its max V so the integration would begin at V=0, t=0
A normal 12th grader can do this
1st view
First
considering potential difference between the terminal of vout ie -iR
shape will be same but lifted up