Hey friends, welcome to the YouTube channel

ALL ABOUT ELECTRONICS. And today we will see the solved problems

based on the magnetically coupled circuits. So, now before we go the problems, let’s briefly

understand what is reactance in the inductive circuit.

As we are going to see this term many times during the problems.

If you already know what is reactance in the inductive circuit, then you can skip this

portion and directly go the example 1 and 2,

The timestamped links are given in the description below.

So, by clicking on them you can directly go the either example 1 or 2.

So reactance is basically a resistance that is offered by the inductor to the AC current.

So, this reactance is very similar to the what is resistance in the resistors in the

DC circuits. And it is denoted by the symbol of X.

And it has the same unit of resistance, that is ohm.

Now, this reactance is basically a ratio of voltage to the current, and it is given by

the expression ωL. Where ω is the angular frequency and L is

the inductance of the inductor. So, as you can see here, as the operating

frequency or inductance of the inductor increases, the value of this reactance increases.

Now, in the case of a resistor, the voltage and the current are in a phase.

But in a purely inductive circuit, whenever the current passing through this inductive

circuit, there is a phase difference between the voltage and current.

So, whenever the current passes through any inductive circuit, the voltage leads the current

by a 90 degree. Or in another term, the current lags the voltage

by 90 degrees. So, if we represent this phase information

in a + jb form, that is a real plus imaginary term, then this

reactance can be given as jωL. Where this jωL represents that the voltage

leads the current by 90 degrees. So, let’s say if you have given one inductor

which is having inductance of 1 H, and if this inductive circuit is operated

at let’s say 5 rad/ sec, then the value of the reactance would be

jωL, that is j5 ohms. Now, if the same circuit is operated at let’s

say 10 rad/sec, then the value of the reactance would be j10

ohm. Now, as we know briefly about the reactance,

so now let’s move on to the examples. So, here in this example, we have given one

circuit which consists two inductors and one resistor.

And this circuit is driven by 50 V AC source. Now, if you see this circuit, in this circuit,

these two inductors are connected in a series and they are mutually coupled.

And we have given that the power that is absorbed in this circuit is 168 W.

And in this circuit, we need to find the mutual inductive reactance between this two inductors.

So, first of all, let’s assume that the current I is flowing through this circuit.

So, in this circuit, these two inductors are connected in series, and they are mutually

coupled to each other. So, let’s say that equivalent reactance is

X. Now, here we don’t know the value of this

mutual reactance and also we don’t know whether this mutual reactance is additive or subtractive.

So, the value of equivalent series reactance would be either j5 + j8 + 2*jXm, if this mutual

reactance is additive. Or it could be a

X=j5 + j8 – 2j*Xm, if this mutual reactance is subtractive.

But in general, we can say that the total equivalent reactance is X.

So, the current I, that is flowing through this circuit can be given as V/(R + jX)

That means, 50 /(10 +jX) So, the magnitude of this current would be

50/sqrt(100+X^2) Or we can write,

I^2=2500/(100 + X^2) And let’s say this equation number 1.

Now, here we have given that the power that is absorbed in this circuit is 168 W.

Now, power always gets absorbed only across the resistive elements.

So, this 168 W of power should get absorbed across this 10 ohm resistor.

So, we know that the power P can be given as, I^2*R

Now, here we have given that power that is absorbed is 168 W and we know that the

value of the resistance is 10 ohm. So, we will get I^2=16.8

And let’s say this is equation number 2. So, now we got equation number 1 and 2.

So, let’s compare this equation number 1 and 2.

So, comparing equation 1 and 2 we will get this equation.

Or we can write it as, 100 + (X^2)=2500/16.8 Now, if we solve this expression, we will

get the value of X as approximately 7. Or we can say that jX=j7 ohm

So, in this way, we found the value of jX is equal to j7 ohm.

So, now if you see this two equations, the addition of j5 and j8 itself is j13 and the

total reactance that we found is j7. So, the total reactance has to be given by

this expression. That is j7=j5 + j8 – 2j*Xm

Or we will get, 2j*Xm=j6 So, we will get a value of this reactance

is 3. So, in this way, we got the value of this

Xm as j3 ohm. And this mutual reactance is subtractive in

nature. So, now let’s see the second example.

So, in this example, we have given two coils which are magnetically coupled to each other.

And we have given that, the inductance between this terminal 1 and 2 is 4 H when this terminal

3 and 4 are open. And when this terminal 3 and 4 are short-circuited

then the value of the inductance that is measured across this terminal 1 and 2 is 3 H.

So, in this example, we need to find the coefficient of coupling.

So, in the earlier videos, we had seen that the coefficient of coupling K can be given

as, K=M/sqrt(L1*L2)

or in this case M/sqrt(Lp*Ls) Now, in this circuit, we had given the value

of this Ls. And first of all to find the coefficient of

coupling, we need to find the value of this Lp and mutual inductance M.

So, to find this value we have given two conditions. So, let’s see the first condition.

So, in the first condition, we have given that, when this terminal 3 and 4 are open

circuited then the value of the inductance that is measured across this terminal 1 and

2 is 4H. That means the value of the inductance that

is measured across this terminal 1 and 2 is 4 H.

Now, whenever this terminal 3 and 4 are open circuited, so no current will flow through

this inductor Ls. So, as no current is flowing through this

coil Ls, so there will not be any coupling with the primary coil or coil Lp.

So, whatever inductance that is measured across this terminal 1 and 2 is, entirely because

of this inductor Lp. So, the value of this Lp has to be 4H.

Now, in the second condition we have given that, whenever you short this terminal 3 and

4, the value of the inductance that is measured across this terminal 1 and 2 is 3H.

So, whenever we short this terminal 3 and 4, the value of the voltage V2 will be zero.

But, now this current I2 will flow through this secondary coil Ls.

So, now if you see, the current I1 and I2 are entering into the dots.

So, the value of the mutual inductance that will be generated in the both the coils will

be positive. So, now let’s apply Kirchhoff’s Voltage Law

or KVL on the primary side. So, we can write

V1=I1*(jωLp) + I2*(jωM) Where, the first term is the voltage that

is induced due to this coil Lp, and the second term represents that voltage that is induced

due to the mutual coupling. Let’s say this equation number 1.

Similarly on the secondary side, as V2 is zero, we will get,

0=I2*(jωLs) + I1*(jωM) Here, also the first term represents the voltage

that is induced due to this coil Ls, While the second term represents the voltage

that is induced due to the mutual coupling. Let’s say this equation number 2.

Now, if we rearrange equation number 2, we will get,

I2=(-M/Ls)*I1 And we know the value of Ls is 2H.

So, we will get I2=(-M/2)*I1 Now, let’s put this value into the equation

number 1. So, we will get

V1=I1*(jωLp), Now, we know the value of Lp is 4H.

So, that is V1=I1*(jω4) + (-M/2)*I1*(jωM)

Now, if we simplify this expression, we will get

V1=jωI1*[4-( M^2)/2] Or we can say that

V1=jωI1*[8-(M^2)]/2 Now, in the second condition, we have given

that the inductance that is measured across terminal

1 and 2 is 3H. So, this term has to be equal to 3H.

That means we can write, [8-(M^2)]/2=3

or we can say that M^2=2 That means, M=sqrt(2)

Now, we know that the coefficient of coupling K can be given as

K=M/sqrt(L1*L2) Or in this case,

K=M/sqrt(Lp*Ls) Now, we know that the value of M is sqrt(2).

The value of Lp is 4H and the value of Ls is 2H.

So, in this way, we will get value of K as 1/2 or 0.5

So, in this way, we found the value of this coefficient of coupling as 0.5.

So, I hope you understood how to solve this two problems.

More examples based on the magnetically coupled circuits are given in the description below.

So, you can try to solve them by yourself. And if you have any question while solving

you can ask me in the comment section below.

For more problems on Magnetically Coupled Circuits, the downloadable google drive link is given in the description.

While solving if you have any questions, do let me know here.

sir doubt power absorption takes place in inductors and capacitors but u said power is absorbed in resistor how.

CORRECTION:

Example 1: Instead of absorbed power, please consider it as the dissipated power in the circuit.

i love your videos thanks for posting

YOU ARE THE BEST.!😘😘GREAT THANKS FOR YOU SIR!

can an english version be made the accent is very annoying

sir how to you take equation j5+j8-2jxm …

and why we cant take quation j5+j8+2jxm….

in example no. 1

nice.

I really like this channel, video was really helpful, thank you

sir can u please make a video on ideal and real transformers?

A brilliant and very clear cut explanation! Thanks!

Please make a video on transient and steady state conditions

How can we determine the coupled equation after getting jx value

i love all about electronics

how is 8-sq(m)/2 =3

Sir very helpfull videos

Why did we consider the value of lp is 4 in 2nd prblm it's probably to be 3 right