a coil of self inductance 50H is joined to the terminals of a battery of e.m.f. 2V through a resistance of 10 ohm and a steady current is flowing through the circuit . if the battery is now disconnected then time in which the current will decay to 1/e of its steady value is
ans: 5 sec
L = 50 H
V = 2 V
r = 10 ohm
After steady state it(inductor) behaves as normal wire:
So the current at steady state becomes = 0.2 Ampere
Now we the expression for decay of current in LR circuit:
i = ioe-rt/L
putting the values we get the value of t as 5 sec
put i= i0/e
V = 2 V
r = 10 ohm
After steady state it(inductor) behaves as normal wire:
So the current at steady state becomes = 0.2 Ampere
Now we the expression for decay of current in LR circuit:
i = ioe-rt/L
putting the values we get the value of t as 5 sec
put i= i0/e