1(2)
you know that,
i = i0(1-e-Rt/L)
Now, the current reaches half of the steady state value....
therefore, put i = i0/2
1/2 = (1 - e-2t/0.3)
2t/0.3 = ln 2
2t = 0.09
t = 0.045 s which is approx. 0.05 s
1) In a series resonant LCR circuit the voltage across R is 100V and R= 1KΩ, C=2μF. The resonant frequency ω is 200 rad s-1. At resonance, the voltage across L is:
a)40 b)250V c)4*10-3V d)2.5*10-2V
2)A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2v. The current reaches half of its steady state value in:
a)0.3s b)0.15s c)0.1s d)0.05s
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4 Answers
11sorry..adarsh
I haven't studied the chapt AC properly... so i cud nt ans ur 1 st q.
49ωL=1ωC
the net resistance will be only due to resistance R
i= VR
V across inductance = i.Xc = i(ωL)=VωRC