Question

Two coils of self-inductance `2mH` and `8 mH` are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil is
A. `10mH`
B. `6mH`
C. `4mH`
D. `16mH`

Answer 1

Correct Answer - C
When the total flux associated with one coil links with the other, i.e., a case of maximum flux linkage, then
`M_(12)=(N_(2)phi_(B)_(2))/(i_(1))` and `M_(21)=(N_(1)phi_(B)_(1))/(i_(2))`
similarly, `L_(1)N_(1)phi_(B)_(1))/(i_(1))` and `L _(2)=(N_(2)phiB_(2))/(i_(2))`
if all the flux of coil `2` links coil `1` and vice versa, then `phi_(B)_(2)=phi_(B)^(1)`
Since, `M_(12)=M_(21)=M`, hence we have
`m_(12)M_(12)=M^(2)=(N_(1)N_(2)phi_(B)_(1)phi_(B)_(2))/(i_(1)i_(2)) =l_(1)L_(2)`
`M_(max)=sqrt(L_(1)L_(2))`
Given, `L_(1)=2mH,L_(2)=8mH`
`M_(max)=sqrt(2xx8)=sqrt(16)=4mH`