ElectroMagnetic InductionHard
Question
In a series L−R growth circuit, if maximum current and maximum induced emf in an inductor of inductance 3mH are 2A and 6V respectively, then the time constant of the circuit is :
Options
A.1 m sec.
B.1/3 m/sec.
C.1/6 m/sec
D.1/2 m/sec
Solution
i = i0(1 − e−t/τ)
|e| = L
= 
e-t/τ ..........(i)
also |e| = emax e-t/τ ........(2)
From (1) and (2)
emax =
⇒ τ = 
= 
= 1m/sec.
|e| = L
= e-t/τ ..........(i)also |e| = emax e-t/τ ........(2)
From (1) and (2)
emax =
⇒ τ = = = 1m/sec.Create a free account to view solution
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