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Question
The variation of inductive reactance (XL) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.
(i) Calculate the self-inductance of the inductor.
(ii) When this inductor is used in series with a capacitor of unknown value and resistor of 10 Ω at 300 s–1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
Solution
(i)
We know that `"X"_"L" = omega"L" and omega = 2pi"f"` where f is frequency in Hz.
So, `"L" = "X"_"L"/(2pif) = (20)/(2pi(100)) = (40)/(2pi(200)) = (60)/(2pi(300)) = 31.84 xx 10^-3 ≈ 32 "mH"`
(ii)
we know that power dissipation is maximum when `"X"_"L" = "X"_"C"`
or `omegaL = 1/(omegaC) or C = 1/(omega^2L`
⇒ `"C" = 1/(4pi^2"f"^2"L") = 1/(4xx3.14xx3.14xx300xx300xx32xx10^-3) = 8.8 mu"F"`
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