Q. 84.4( 20 Votes )
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self - inductance of the circuit.
Answer :
Given: Initial current I1 = 5.0 A
Final current I2 = 0.0 A
Time = 0.1 seconds
Average e.m.f = 200 V
Change in the current can be calculated as:
dI = I1 - I2 = 5.0A – 0.0A
dI = 5 A
The self-conductance of the circuit is calculated using the formula:
or
substituting the values, we get
∴ L = 4 Henry
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PREVIOUSA horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.(a) What is the instantaneous value of the emf induced in the wire?(b) What is the direction of the emf?(c) Which end of the wire is at the higher electrical potential?NEXTA pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
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