Question 108 of 153

Two coils of self inductance ${L}_{1}$ and ${L}_{2}$ are connected in series combination having mutual inductance of the coils as $M$. The equivalent self inductance of the combination will be

A$\dfrac{1}{{L}_{1}}+\dfrac{1}{{L}_{2}}+\dfrac{1}{M}$B${L}_{1}+{L}_{2}-2M$C${L}_{1}+{L}_{2}+M$D${L}_{1}+{L}_{2}+2M$

Show answer and solution▾

Answer: (B) ${L}_{1}+{L}_{2}-2M$

Here self inductances are in series but their mutual inductances are linked oppositely.
As the inductors connected are in series opposition, hence magnetic flux will be opposing each other. Therefore,
${L}_{eq}={L}_{1}+{L}_{2}-2M$