Two coils of self-inductance L1 and L2 are connected in series combination having mutual inductance of the coils as M. The equivalent self-inductance of the combination will be:
\(\frac{1}{L1 }\) + \(\frac{1}{L2}\) + \(\frac{1}{M}\)
- L1 + L2 + M
- L1 + L2 + 2M
- L1 + L2 – 2M
The Correct Option is D
Solution and Explanation
Self-inductances are in series but their mutual inductances are linked oppositely so equivalent self-inductance
L = L1 + L2 – M – M
L = L1 + L2 – 2M
\(\therefore,\) The correct option is (D): \(L_1 + L_2 – 2M\)
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JEE Main Notification
GMC Azamgarh Admission 2025: Dates, Courses, Fees, Eligibility, SelectJuly 22, 2025Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
- When we place the conductor in a changing magnetic field.
- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
- t = time
Applications of Electromagnetic Induction
- Electromagnetic induction in AC generator
- Electrical Transformers
- Magnetic Flow Meter