0
\$\begingroup\$

enter image description here

In the above diagram the circuit is in Series Resonance. Now we are asked to find the coupling coefficient K between coils L1 and L2. For which i first calculated L(equivalent) as follows :

Leq = L1 + L2 + 2M ( Dots are in aiding position)
Leq = 12 + 3 + 2M (Assuming w = 1)
Leq = 15 + 2M (M denotes mutual inductance between two coils)

So the circuit becomes

enter image description here

Since The circuit is in resonace the XL should equal to Xc

  Xc = XL
-jw18 = jw(15+2M)
 -18  = 15 + 2M ( w = 1)
  -16.5 = M

Now, To find coupling coefficient K, we can use the follow formula

K = M / sqrt(L1*L2)
K = -16.5/ sqrt(12*3) = -2.75

Which is not possible, Since Range of k should be 0 < K < 1. Where am i wrong ? Please Help...

\$\endgroup\$
1
\$\begingroup\$

Hint - if the two inductors were perfectly coupled (k = 1) then you would get a total reactance of j27 ohms. Are you able to see that? If you are then clearly, to obtain a total reactance of j18 ohms k must be less than 1.

How did I get j27 ohms - it's a sanity check - reactance or inductance is proportional to turns squared and if coupling is 100% then the turns for L1 and L2 are perfectly coupled and can be regarded as being on the same former so, new impedance is \$j(\sqrt{12} + \sqrt{3})^2\$ = j27 ohms.

I think the mistake I see is that you equate the reactance to -j18 when in fact M should equal j3.

\$\endgroup\$
  • \$\begingroup\$ @MayankPal Did this answer help you? \$\endgroup\$ – Andy aka Oct 25 '16 at 16:57
0
\$\begingroup\$

Here equivalent Inductance is 15+2M but in resonance condition capacitive reactance equal to inductive reactance so here 18=15+2M then M value becomes 1.5 from this K=0.25

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.