In the last classes we have studied about

the application of op-amp and we have seen application of op-amp in various arithmetic

circuits like logarithmic, exponential as well as precision half wave rectifiers, etc. Today we will study about the application

of op-amp in control sources. What is a control source? Control source means we can control the output

with a control input. That is suppose we want to vary the output

current of a circuit by varying the input voltage or suppose we want to vary the output

voltage by varying the input voltage or suppose we want to vary the output current by varying

the input current; so the different examples of control sources are there. Here the source means a voltage or a current

source and that voltage or current source which we will get are in fact obtained by

input voltage or current sources. A typical example which we earlier studied

was that of a transistor. If we consider a transistor, we get at the

output a control source or a dependent current source because if we consider a common emitter

transistor, the output current is the collector current and that collector current IC is equal

to beta times of the base current. This example is nothing but that of a current

controlled current source. Here we are controlling the output current

IC by controlling the input current IV. Similarly using op-amp also we can have the

control sources and that we will discuss today and the control sources are having various

configurations like voltage controlled voltage source. It is called VCVS, then voltage controlled

current source, VCCS it is in short called and current controlled voltage source that

is CCVS or current controlled current source that is CCCS. These different types of control sources we

can have by using op-amp. Let us take a schematic diagram of a voltage

controlled voltage source or VCVS. If we consider a block of a VCVS, the input

is Vi that we are giving. We have to give an input and as it is controlled

by voltage, input will be a voltage and that is Vi and the voltage source at the output

or the control source which we will get at the output is also voltage and that is VO. The main thing here is that VO is controlled

by Vi. That is we will get VO, which is controlled

or affected by Vi and that is equal to k times Vi we are writing in order to express that

relationship between output voltage and input voltage. Here this k which is a constant, it will relate

the input voltage and output voltage. If we vary this input voltage, output voltage

here will accordingly vary. This is a block diagram for representation

of an ideal voltage controlled voltage source. Here we are using the term ideal because we

are not considering the practical parameters which exist. For example we know in an op-amp there will

be input impedance, there will be output impedance. But in this ideal consideration, we are not

considering those or we are simply ignoring those and an ideal op-amp consideration still

we are using ignoring the input impedance and output impedance. Let us consider a practical circuit. This circuit is a very familiar circuit. We used it when we were discussing about the

inverting op-amp; inverting amplifier when we discussed earlier, the same circuit configuration

we have used. Input voltage Vi is having an input resistance

which is connected that is R1 and the feedback resistance is Rf and the non-inverting terminal

is grounded. What will be the output voltage VO and from

our earlier study we already know VO equal to minus Rf by R1 into Vi and Rf by R1 for

a circuit once we design it, suppose we are not varying it, we can name it as k. k being

the ratio between Rf and R1 that is the feedback resistance and the input resistance and that

k is relating the input voltage with output voltage. That means we are controlling the output voltage

VO with an input voltage Vi. The output and input both are voltage sources. This is an example of a controlled voltage

source and this voltage source which is controlled is by another voltage source only at the input. Similarly the non-inverting op-amp example

which we are discussing earlier is also an example of voltage controlled voltage source. Because if we see in the circuit we are connecting

a voltage Vi at the non-inverting terminal and there is a resistance RC. This RC resistance is there but it is not

going to affect the overall expression of the output voltage VO. Because if we want to find out what is the

output voltage VO, VO is here and if we see the side where Vi is connected, this RC is

there, but there no current into the op-amp. If we name this current, which we have flowing

through this Rf, the same current will flow through this R1 and this example and the non-inverting

terminal op-amp which we were discussing earlier does not have any difference in analysis because

this presence of RC is not going to affect anything. What will be the output voltage VO? VO is again 1 plus Rf by R1 into Vi only. This example is also an example of a voltage

controlled voltage source. The output voltage VO is controlled by the

input voltage Vi. Another control source is voltage controlled

current source and in short it is known as VCCS. As the name suggests voltage controlled current

source means the output source will be the current source, but it will be controlled

by an input voltage. Here is the circuit of a voltage controlled

current source. We are having a voltage at the input which

is Vi and if we now denote the current flowing in this resistance R, this is Ii. We are naming it Ii. The same current will also flow through the

resistance RL because there cannot be any flow in this op-amp. Current flow cannot enter in the op-amp; whatever

current is flowing from this Vi must flow through this RL. So the current IO is same as Ii and what is

Ii? This is nothing but Vi by R only because we

will go by this loop. So Vi minus zero by R that is Ii and it can

be written as 1 by R into Vi; this 1 by R, let us name by k. So what do we get at the output? Here, the current IO we are getting from Vi

with a relation that is k times of Vi or k is equal to 1 by R or 1 by R times Vi is equal

to IO. If we vary this Vi, the output current will

also vary. Another example of voltage controlled current

source can be this example where we are having a Vi at the non-inverting terminal and we

are having resistances R and RL connected in this manner, RL being the feedback resistance. This circuit and the earlier circuit (Refer

Slide Time: 9:32) have a difference. Where is the difference? Here we are having a RL in this case which

is floating. Floating means it is not grounded. If you see both the sides, this side is connected

to R and this side is connected which is the output voltage VO. This point is not ground; so, this is called

a floating resistance or floating load resistance RL. But here in this circuit when the load is

floating we will have this control source, which is same as the example which we were

considering earlier which is also nothing but a non-inverting op-amp circuit. We will discuss a circuit which is having

a floating resistance and a circuit which will have a non floating resistance; in both

the cases we can use actually a voltage controlled current source. In this circuit we are having the output current

IO which is effected or which is controlled by the input voltage source Vi. The voltage controlled current source that

we are considering right now, finds application in electronic voltmeters. If we consider electronic voltmeters circuit,

electronic voltmeter is basically a voltmeter which should be able to measure even very

small voltages and if we want to find the measurement for a voltage we must have a considerable

deflection in the meter and we may also want that a full scale deflection should occur

for a specified voltage which is to be measured. In those cases, we will use a voltage controlled

current source like the example which we discussed just now. Here these examples show a moving coil meter,

the basic component in an electronic voltmeter and this coil resistance is RL. We want to measure a voltage which is V. We are using it with an op-amp that means

we are not using this electronic voltmeter alone. This moving coil meter, which we are having

for measurement we are not simply using it and measuring any voltage, we are combining

it with an op-amp. The reason for this combination with an op-amp

is that we want to have a specified deflection in this meter with respect to a particular

voltage being applied at the input. Suppose we are measuring a millivolt range

of a voltage and we want to have a full scale deflection in the moving coil meter which

will be basically dependent upon the current flowing through it. The current which will flow if we denote by

I that is the current which is flowing through this moving coil meter having the resistance

RL. This current should produce that much of deflection. Suppose if we want to have a deflection of

100 milliampere for a millivoltage range which is specified then how we can achieve this,

is dependent upon how much resistance we will connect at R, because this circuit and the

circuit (Refer Slide Time: 11:30) which you discussed which is voltage controlled current

source are typically same because there we were using a feedback resistance which is

RL. This RL is now the resistance of the coil

resistance of the moving coil meter. Here the voltage which is to be measured is

given at the input, non-inverting terminal. The current which is flowing through this

coil or the moving coil meter is I. So this current is equal to, from the earlier

consideration we have seen, Vi by R. R is the key element which will decide the

current and in order to have a specified value of current for a specified amount of voltage

at the input, we must connect that R which will give our requirement. This is a very important example of application

of voltage controlled current source and here one thing that is achieved by combining it

with an op-amp is that the voltmeter input resistance is almost infinite we are getting

and that is very useful. Also another factor is that the scaling factor

will only depend upon the resistances because here whatever the output current must be that

is decided by the resistance and that is why we use this type of VCCS in electronic voltmeters

where we can even measure very small millivolt range of the voltage very accurately. If we consider this example a voltage controlled

current source, here we are having a same VCCS that is voltage controlled current source

which is the similar example as the earlier case. But there is a difference here. In this case we see that the load resistance

which is connected is grounded. In the earlier cases, here (Refer Slide Time:

11:30) as well as here also (Refer Slide Time: 9:32), the load resistance is not grounded,

but here the load resistance is grounded. This is also an example of a voltage controlled

current source. The output current IO is the current of interest. That is we want to find out what will be this

IO? But our input is a voltage which is Vi. In order to know what is IO let us first assume

one thing, before finally analyzing the circuit, that R3 by R1 the ratio between these two

resistances is equal to R4 by R2 that is maintained and if this resistance ratio is maintained

then we can show that IO which will be finally flowing at the output is equal to minus Vi

by R2, Vi being the input voltage. We can show that the current at the output

IO is equal to minus Vi by R2 if R3 by R1 is equal to R4 by R2. Let us do this example. We will try to show that IO is equal to minus

Vi by R2. Drawing this circuit again, let us name the

currents which are flowing. The current which is flowing through R1 is

i1 and the same current i1 will also flow through R3 that is also i1 because there cannot

be any flow of current into the op-amp which is the condition or which is the assumption

which we have been following, following the ideal op-amp consideration. What will be the current IO, we have to find

and for that let us name one node voltage V1 which is here. Let us name this node voltage as V1. If this is V1, this node voltage is also V1

because of this op-amp being ideal. This voltage will be also V1 that means this

voltage is also V1. This point and this point are same. We now apply Kirchhoff’s current law, KCL

at this node V1. One current is IO, which is outgoing and let

us also assume the current which is flowing through R2 that is also outgoing and the current

which is flowing through RO is say incoming. This again is our assumption. You can also assume it as outgoing or this

is incoming as you wish because ultimately we will get same expression only if we follow

Kirchhoff’s current law. Let us name it as the current here is say

i2. This is i3. If we apply Kirchhoff’s law at V1, we get

the sum of the incoming current is equal to the sum of the outgoing currents. Here the outgoing currents are i2 plus IO

that is equal to the incoming current i3. What is i2? i2 current is nothing but V1 by R2 and what

is i3? We can find out i3 because i3 is nothing but

the difference in potential between this point and this point which is VO minus V1 divided

by R4. Using this result that is V1 by R2 plus IO

these are the outgoing currents, we have denoted like this and equal to incoming current which

is equal to VO minus V1 by R4; that is the Kirchhoff’s current law being applied at

node V1. You will further simplify because we know

what is V1? If we find out V1 from this input side, what

is V1? This voltage is nothing but Vi minus this

drop which is i1 into R1. So V1 equal to Vi minus i1 into R1; we will

use that here. Instead of writing V1, we will write Vi minus

i1 R1 by R2 plus let us say keep IO, as it is because we want to find out IO and from

the figure what is VO? We can write VO in terms of V1 as V1 minus

i1 R3; this drop when you subtract from V1 that will be VO. That is VO can be written as V1 minus i1 into

R3 by R4 minus V1 by R4 let us separate. We are keeping V1 in all the expression and

if we now separate out some of the terms, Vi by R2 minus i1 into R1 by R2 plus IO equal

to again V1 can be written as Vi minus i1 R1 by R minus i1 R3 by i1 R3 by R4 minus again

V1 we can write Vi minus i1 R1 be careful about these signs plus and minus. This is the expression. Now we write as Vi by R2 plus IO and transfer

all the other terms to the right side. If I transfer these terms to the right side

that means i1 into R1 by R2 will be positive plus what we will have here is Vi by R4 minus

i1 R1 by R4 minus i1 R3 by R4 minus Vi by R4 plus i1 R1 by R4. If you look into this expression, there will

be some terms which will cancel out; this one and this one Vi by R4 and this Vi by R4

minus will cancel out and i1 R1 by R4 and minus i1 R1 by R4 will cancel out. So we are left with these two terms in the

right side, which is equal to i1 into R1 by R2 minus R3 by R4 and that is equal to left

side which is Vi R2 plus IO. Now look into the right side because we are

using an assumption which is given to us that is R3 by R1 is equal to R4 by R2. We will use that; R3 by R4 is equal to R1

by R2. Using this expression we will now have these

two cancel out, so these two will cancel out. We will get here zero that means Vi by R2

plus IO equal to zero. If the right hand side of this Vi by R2 plus

IO is zero that means we get IO equal to minus Vi by R2. That means we can show that the controlled

source current or current source which we will get at the output which is IO that is

equal to minus 1 by R2 into input voltage Vi. Our result is now obtained which you have

been asked to get by using this law or by using this relation actually. This example was an example of a voltage controlled

current source. Another source which is a controlled source

is current controlled voltage source. So far we have discussed about voltage controlled

voltage source as well as voltage controlled current source. Now we will discuss current controlled voltage

source and current controlled current source. Current controlled voltage source or CCVS,

as it is named, is having this diagram which is an ideal block diagram form of a current

controlled voltage source. Here the input is a current. Ii is current and at the output we are getting

a voltage which is VO and VO is a voltage which will be controlled by the input current

Ii according to the law k times of Ii; k is actually a constant. For example we have this circuit where we

are having an input current Ii and the voltage at the output if we want to find out what

is this VO? VO is found out by flowing in this loop starting

from ground which is zero minus IO into RL is equal to VO or IO we can just write as

Ii. Because IO is equal to Ii there cannot be

any current flow in the op-amp. So zero minus Ii into RL is equal to VO. So that means VO is equal to minus Ii into

RL. Here the k which we are writing here is this

resistance RL. This example is a very simple example of the

current controlled voltage source. Basically in the example which we have shown

here, input is having a current source. But if we think about a current source, basically

a current source is having actually an infinite resistance ideally, in parallel, and as larger

value that parallel resistance has, it is more and more towards an ideal current source. Because we know that along with the voltage

and current there will be series resistance in voltage source and current source will

have a parallel resistance. An ideal voltage source is when the series

resistance is zero and an ideal current source is when the parallel resistance is infinite. But practically we never get an infinite parallel

resistance with a current source and that is why even if we want that the input current

should pass on to the load wholly, that means 100% of the current available from the current

source must be available for the input circuit, that does not happen because a part is lost

as the current through the parallel resistance which is along with it. This is a current source. Basically a practical current source will

have this resistance, RS. We apply directly a current source in a circuit. In this circuit whatever we have shown, it

is a current source just applied in this circuit; but then the current source is having a resistance

that was not shown in this case. But because of this fact that it has a resistance

which is parallel to it and which is to be avoided and because we want that the resistance

should not draw any current so that the whole source current must be available in the circuit,

we are combining it with an op-amp. What will we gain? We now see what part of the current will flow

through this resistance? In this resistance, this point is the negative

or inverting terminal of the op-amp and its positive is grounded. We are having an op-amp. These two points are having the same ground

voltage. This voltage is also ground voltage and at

the other point of this source resistance, this is also ground. There cannot be any flow of current through

this resistance RS because both the points are at ground because of the connection through

an op-amp. Our aim that there should not be any current

flow in the parallel resistance of the current source is being achieved and so the current

will be flowing wholly through this R and so the output voltage we will get minus R

into IS. This is an example of the current controlled

voltage source because we are getting an output voltage VO which is controlled by the current

source IS. Now current controlled current source is another

example of control source driven by an input current and here the output is a current. Earlier case we got a current controlled voltage

source but here the current controlled current source we are getting. This is the ideal block diagram representation

of this current controlled current source. This is the model of the current controlled

current source. Here this is the dependent current source

at the output which we obtain from the input by this relation that IO is equal to k times

of Ii. Let us now see the practical implementation

of a current controlled current source. This is an example of a current controlled

current source. Here we are having an input current and we

have to find out what will be the output current IO? There is this op-amp and positive or non-inverting

terminal is grounded and we are having resistive network having the resistances R1, R2 and

RL. The current if we denote at the input as I1,

this I1 will flow through R1 as there cannot be a current in the op-amp and at this point,

one part of the current will be say I2 and the other current is IO. So I1 is basically IO plus I2. So IO is I1 minus I2. This current I2 we can find out. If I name this voltage as V, the current I2

is nothing but V by R2; this is I2. Now what is V? Suppose we use this loop; this current is

flowing. This point is plus, this point is minus. So V plus I1 R1 is equal to zero; because

I am coming from this point downwards, minus plus is raising voltage, so V plus I1 into

R1 is equal to zero. This is ground. From here I get what is V? V is equal to minus I1 R1. I am replacing this minus I1 R1 here. In place of I2 I am writing minus I1 R1 divided

by R2. That is the expression for I2. Simplifying further, taking I1 common I get

1 plus R1 by R2. So IO is equal to 1 plus R1 by R2 into I1. This can be written as k. This is the ratio R1 by R2 plus 1. This is a constant. That means we get this expression of this

IO as k times of I1 and this is the form of a controlled source which is a current source

driven by input current. This is an example of a current controlled

current source, CCCS. We are controlling the output current by an

input current. Example of control sources we have taken up

today. We have seen different forms of control sources;

voltage controlled voltage source, voltage controlled current source, current controlled

voltage source and current controlled current source. Let us now try to solve one or two examples

which are based on these op-amp circuits. Using that ideal op-amp consideration, now

let us find out a voltage VO given in a circuit like this. We have an op-amp and we are having a source

applied to it say minus plus, minus plus 2 volt and this resistance, if this one is 20

kilo ohm, VO is to be found out and the positive terminal is grounded but the point here is

connected to this positive terminal and this resistance is 10 kilo ohm. Note this circuit using this op-amp. We have to find what will be the value of

VO? In all analyses we use ideal op-amp consideration;

that means we will use that property of ideal op-amp. In order to find out VO, let us name this

node voltage as VX. What is VX? If we come down from this point to ground

VX minus 2 volt is equal to zero. So VX is equal to 2 volt and let us name the

current as i, which flows through 20 k. I am showing the direction like this; intuitively

I am assuming that this is flowing to ground, so it will be in this direction. But there is no hard and fast rule. We can take the direction from left to right

also; that will not harm. We now find out what is I? I, current can be found out if we come from

the VX to ground through this resistance. The current i, which is flowing will not go

through this part because then it will have to enter the op-amp. So it will flow down like this. This current i and this current i, which flows

through k, 10 k is same. So now I can write VX minus 10 k into i minus

zero equal to zero. So what is i can be found out; VX by 10 k

ohm and we know VX is 2 volt. So 2 volt by 10 kilo ohm is the value of this

current. That means it is 0.2 milliampere. We got the current i is equal to 0.2 milliampere. We are interested in finding out VO. How we can find out VO? VO can be found out if we flow from this point

VO minus 20 into i minus 2 volt is equal to zero. The voltage VO can be found out using this

expression or KVL; VO minus i into 20 this drop minus 2 volt to ground if we come. VO is equal to 2 plus i into 20; 20 is in

kilo ohm and i we have found out in milliampere. So it will be in volt only; 0.2 milliampere

into 20, so 2 plus 4 which is 6 volt is the voltage VO. This is an example of a simple circuit. Similarly we can proceed to find out actually

different parameters in a circuit. Another example let us do to find out the

voltage in this circuit. If we have a circuit having say a resistance

is connected here which is 10 kilo ohm, another resistance is here which is also 10 kilo ohm

then this resistance is 20 kilo ohm. There is a source or a battery 2 volt. This point is ground and another source is

in the opposite direction of polarity. This is 3 volt. Another resistance is there in the feedback

part which is having 20 k ohm. We have to find out what is this voltage VO? To solve this example, let us name the node

voltages as v1 here. If this is v1, this is also v1 according to

the ideal op-amp assumptions or ideal op-amp properties and let us name the current also. This current is i let us name it and the current

which is flowing through this 10 k ohm must be also equal to i. We have to find out what is VO and for that

let us first find out what is this node voltage V1? If you see

this circuit, we want to find out the voltage at this point. The other point is grounded. This is a circuit where we can find out the

voltage V1 simply by following the voltage division law. So V1 is equal to 2 into 20 by 20 plus 10;

kilo ohm I am omitting because all are in the same unit. So we get 2 into 20 by 30. That means we are getting a 4 by 3 volt at

this node which is same as this voltage. V1 is equal to 4 by 3 volt. We can now apply the Kirchhoff’s current

law at this node V1. If we apply Kirchhoff’s current law, at

this node V1 the current incoming is i, which is equal to VO minus V1 divided by 20 k and

that is equal to the current outgoing which is equal to V1 minus -3 because this polarity

we have to be careful. This is minus, this is plus, so if I find

out the potential difference between these two points, this point is grounded and this

point, V1 minus this voltage but this voltage is -3; so V1 plus 3 divided by 10 k that is

the outgoing current. This expression can be applied here at V1

of the Kirchhoff’s current law at node V1. KCL, Kirchhoff’s current law we are applying

at node, which has a voltage V1. So that gives us the incoming current i is

equal to the outgoing current, same current but we will find out this expression from

the node voltages VO minus V1 by 20 k. Omitting this k because both sides will have

the same unit equal to as I have said V1 minus -3 that is V1 plus 3 by 10 and we already

have known what is the value of V1. V1 is equal to 4 by 3 volt. So this is 2; so from this relation we get

VO minus V1 equal to 2 times V1 plus 6 or what is V1 can be found out? VO is equal to 3 times V1 plus 6. So replacing this value of V1 by 4 by 3 what

we get at the output is 10 volt. The output voltage VO is 10 volt that we have

got here in this example following the principal of ideal op-amp. In another example let us try to find out

the power in a circuit. For example we have a circuit having an op-amp

here. We are having this battery here minus plus

minus plus 4 volt. 4 volt battery is applied in the feedback

part and we are having in the non-inverting terminal, a resistance which is 30 kilo ohm

and another resistance is connected which is 20 kilo ohm and here there is a resistance

which is 10 kilo ohm to ground. This is the circuit. We want to find out the output voltage. Find VO as well as the power supplied by the

4 volt source. In order to find out the power supplied by

the 4 volt source, we must also know the current flowing through it. Because power is equal to voltage into current,

we must know the current which flows. Let us name it by i. So we will have to find out the current through

it because then only we can find out what is the power supplied by the 4 volt source. Power supplied by the 4 volt source will be

4 into i. So i we need to know. In order to solve this following similarly

let us name this node voltage here as v1 which is same as this point voltage which is V1

only; these two voltages are same. We now find out what is? We can find out V1 in one way just by finding

out this voltage which is that portion of the voltage available from VO. We do not know VO but then we will be solving

following this relation. What is V1? VO into 30 k divided by 30 plus 20 k; because

if we consider this as a source this is ground, this is the part. The voltage here into this resistance by this

resistance plus this resistance; this is the voltage division principal we are applying. We are interested to know this voltage. We are finding out VO into 30 by 30 plus 20;

so 3 by 5, 30 by 50 is 3 by 5. So 3 by 5 VO is equal to V1. We have found this voltage, here as well here. This is the same voltage 3 by 5 into VO and

so this can be written in another way VO minus 4 volt is equal to again V1 that is equal

to VO minus 4 volt. Using these two relations we can find what

is VO? So VO minus 4 is equal to 3 by 5 VO or we

can find what is VO? VO is equal to VO into 1 minus 3 by 5 equal

to 4 VO into 2 by 5 is equal to 4. So VO equal to 10 volt. We have known 10 volt is VO. If we know VO, then V1 we can find out. V1 is nothing but VO minus 4; the voltage

here is VO minus 4. 10 minus 4 is 6 volt. We know what is V1 and so we can now find

out what is the current flowing i? i, this current is nothing but the same current;

this current is same current i, so V1 by 10 k. V1 is 6 volt so i is equal to V1 by 10 k;

6 by 10 is 0.6 milliampere which is i. The current which flows here is 0.6 milliampere. 0.6 milliampere is the current which is flowing;

i is equal to 0.6 milliampere, the current we know. We know the voltage which is 4 volt; so, we

know the power supplied by the battery is 4 volt. The power supplied by the 4 volt source equal

to 4 volt multiplied by i and 4 volt multiplied by i, we have found out to be 0.6 milliampere

and that gives us 4 into 0.6 which is 2.4 milliwatts; volt into ampere is watts. So miiliampere into volt is milliwatt. That means we have found out the power supplied

by the 4 volt source which is 2.4 milliwatt and for that we have to find out the current

flowing through that circuit or we have to know the current supplied by the source 4

volt actually. In this example we have seen that power can

also be found out if we know the current as well as the voltage. These are some of the examples which we have

solved today by following the same analysis of op-amp being ideal. In the whole analysis or in all the examples

we have solved till now we are following only the ideal op-amp consideration. In order to solve such problems we have to

know the ideal op-amp consideration only and then we can apply Kirchhoff’s voltage law

as well as current law and we can find out the required quantities whether it is voltage

or current in the circuit. In today’s discussions we have seen the

control sources being built up using op-amp and the four different variations of the control

sources we have studied today. These four variations of control sources are

voltage controlled voltage source, voltage controlled current source, current controlled

voltage source and current controlled current source and these are extensively used. Mainly in instrumentation circuits also you

will find applications of these control sources and also we have seen how we can solve numerical

examples using ideal op-amp considerations and the KVL, KCL, etc to arrive at the required

solutions.

show you tube

With due regards … Its really tough to understand when the quality of the aid used is so poor. When the professor is pointing on certain points on the circuit its difficult to understand which point she is talking about as the pointer is invisible. I guess the video not taking the output from the device she is working rather its capturing the the screen by a camera.

in the problem given in the lecture i don't think the solution was right because negative terminal of an opamp is usually considered as a virtual ground as positive is grounded and if it is so why is it considered the same current in the upper loop too.

if any one gets the logic please help me. For all those thanks in advance.

The below comment corresponds to first example in the lecture.