A 200 resistance, a 0.1 F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
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Answer:
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Answer:
70.4 H
Explanation:
Given, C = 0.1μF = 10^−7 F
Resonant frequency =60Hz
We know that, at resonance
ω0=1/√LC
⇒ 2πf0=1/√LC
⇒L=1 / 4π^2f^20 C
by putting values
L=1 / 4×(3.14)^2×60^2×10^−7
= 10^7 / 4×(3.14)^2×3600
⇒L=70.43 H