Two coils of self-inductance $L_{1}$ and $L_{2}$ are connected in series. If the mutual inductance of the coils is $M$,the equivalent self-inductance of the combination shown in the figure will be:

  • A
    $\frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{M}$
  • B
    $L_{1} + L_{2} + M$
  • C
    $L_{1} + L_{2} + 2M$
  • D
    $L_{1} + L_{2} - 2M$

Explore More

Similar Questions

$A$ coil of inductance $L$ is divided into $6$ equal parts. All these parts are connected in parallel. The resultant inductance of this combination is

Two pure inductors each of self inductance $L$ are connected in parallel but are well separated from each other. The total inductance is

The equivalent inductance between $A$ and $B$ is equal to

Two identical coils of inductance $L$ joined in series are placed very close to each other such that the winding direction of one coil is exactly opposite to that of the other. The net inductance is

Three inductances are connected as shown in the figure. The equivalent inductance between $A$ and $B$ is (in $H$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo