Question

Two inductors of inductance 9H and 4H are couples in series. The separation between them and relative orientation is fixed and once they are connected ine configuration A and then in configuration B as below. Is the maximum and minimum equivalent inductance of the combination is found to be 16H and 3H, the coupling constant is

Answer 1

Answer:

Explanation:

For two inductors : $L_1$ and $L_2$ connected in series,

(i) If the mutual inductor supports self flux,

   the flux through first inductor is $\phi_1=L_1i+Mi$ and for second inductor           $\phi_2=L_2i+Mi$

Now the net voltage will be sum of voltages over both inductors:

$\epsilon_{net} = \epsilon_1+\epsilon_2$

$L_{eq} \frac{di}{dt}=(L_1+L_2+2M)\frac{di}{dt}\\L_{eq_{max}}=L_1+L_2+2M$

Similarly the minimum can be found by $\phi_1=L_1i-Mi , \ \phi_2=L_2i-Mi$

$L_{eq_{min}}=L_1+L_2-2M$

Now, using the definition of Mutual inductance using the coupling constant (k):

$M=k\sqrt{L_1L_2}$

We have:

$16=9+4+6k\\k=\frac{1}{2}$