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Question
L-C-R series circuit contains a resistance of 10 Ω and self-inductance 0.4 H connected in series with variable capacitor across 60 V and 50 Hz supply. The value of capacity at resonance will be π2 = 10.
Options
25 μF
26 μF
22 μF
24 μF
MCQ
Solution
25 μF
Explanation:
f = `1/(2pi sqrt"LC")`
`therefore "C" = 1/(4pi^2 "Lf"^2) = 25 xx 10^-6 "F"`
= 25 μF
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Power in AC Circuit
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