NEET-XII-Physics
39: Alternating Current
- #9A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for ω = 100 s-1, 500 s-1, 1000 s-1.Ans : Given:
Inductance of the coil, L = 5.0 mH = 0.005 H
(a) At ω = 100 s-1:
Reactance of coil `` \left({X}_{L}\right)`` is given by,
XL = `` \omega L``
Here, `` \omega `` = angular frequency
`` \therefore `` XL = 100 `` \times `` 0.005 = 0.5 `` \,\mathrm{\,\Omega \,}``
Peak current, I0 = `` \frac{10}{0.5}=20\,\mathrm{\,A\,}``
(b) At ω = 500 s-1:
`` \,\mathrm{\,Reactance\,},{X}_{\,\mathrm{\,L\,}}=500\times \frac{5}{1000}``
`` =2.5\,\mathrm{\,\Omega \,}``
`` ``
Peak current, I0 = `` \frac{10}{2.5}=4\,\mathrm{\,A\,}``
(c) ω = 1000 s-1:
Reactance, XL = 1000 `` \times 0.005`` = 5 `` \,\mathrm{\,\Omega \,}``
Peak current, I0 = `` \frac{10}{5}`` = 2 A
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