Advertisements
Advertisements
प्रश्न
In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf induced in the coil.
Advertisements
उत्तर
Self inductance, L=5 mH=5×10−3 HChange in current,
dI=(1−4)=−3 A
Change in time, `dt=30 ms=30×10−3 s`
The emf induced in the coil is given by ` e=−L(dI)/(dt)`
`⇒e=((−5xx10^(−3) )xx(−3))/(30×10^(−3))`
`⇒e=0.5 V`
APPEARS IN
संबंधित प्रश्न
Write the SI unit and dimention of of co-efficient of self induction
Obtain an expression for the self-inductance of a solenoid.
Consider a solenoid carrying supplied by a source with a constant emf containing iron core inside it. When the core is pulled out of the solenoid, the change in current will ______.
A toroidal solenoid with air core has an average radius 'R', number of turns 'N' and area of cross-section 'A'. The self-inductance of the solenoid is (Neglect the field variation cross-section of the toroid)
The frequency of γ-rays, X-rays and ultraviolet rays are a, b and c respectively. Then, ______.
A coil is wound on a frame of rectangular cross-section. If all the linear dimensions of the frame are increased by a factor 2 and the number of turns per unit length of the coil remains the same, self-inductance of the coil increases by a factor of ______.
The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as ______.
Obtain an expression for the self inductance of a solenoid.
The current flowing through an inductor of self-inductance L is continuously increasing at constant rate. The variation of induced e.m.f. (e) verses dI/dt is shown graphically by figure
