# Solved Problems on Magnetically Coupled Circuits

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.

## 17 thoughts on “Solved Problems on Magnetically Coupled Circuits”

1. ALL ABOUT ELECTRONICS says:

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

2. raj yadav says:

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

3. ALL ABOUT ELECTRONICS says:

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

4. Sparkling Wish says:

i love your videos thanks for posting

5. ravi indra says:

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

6. Samuel Derbyshire says:

can an english version be made the accent is very annoying

7. pranjal Srivastava says:

sir how to you take equation j5+j8-2jxm …
and why we cant take quation j5+j8+2jxm….
in example no. 1

8. Rudrapratap Bansidhar says:

nice.

9. Apoorva 316 says:

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

10. Abhas das says:

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

11. Bent Hestad says:

A brilliant and very clear cut explanation! Thanks!

12. abhijith m says:

13. Devasish Bhagawati says:

How can we determine the coupled equation after getting jx value

14. pratyush harsh says:

15. shams ul haq says:
16. Rohan kademani says:
17. Chandana Priya says: