Question
Download Solution PDFWhat will be the self-inductance of the coil if an EMF of 10 V is induced in it when the current flowing through it changes at the rate of 5 A/sec?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Self-Induction
- Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
- As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
- This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
Induced e.m.f can be given as
\(⇒ V_{l}=-L\frac{dI}{dt}\) -----(1)
Where
VL = induced voltage in volts
N = self-inductance of the coil
\(\frac{dI}{dt}=\) rate of change of current in ampere/second
Calculation:
Given: di/dt = 5 A/sec, V = 10 volt
From equation 1,
\(\Rightarrow 10=-L\times 5\)
The negative sign indicates that induced emf (e) opposes the change of flux.
⇒ L = 2 H
Last updated on May 29, 2025
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