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प्रश्न
A capacitor of 100 μF, a coil of resistance 50Ω, and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
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उत्तर
Data: C = 100 µf =100 x 10-6 F = 10-4 F,
R = 50Ω, L = 0.5H, f = 50 Hz, Vrms = 110V
∴ ωL = 2πfL = 2(3.142)(50)(0.5) = 157.1Ω
and `1/(omega"C") = 1/(2pi"fC") = 1/(2(3.142)(50)(10^-4))`
`= 100/3.142 = 31.83 Omega`
`"Z"^2 = "R"^2 + (omega"L" - 1/(omega"C"))^2 = (50)^2 + (157.1 - 31.83)^2`
= 2500 + 15700 = 18200 Ω2
∴ Impedance, Z = `sqrt 18200` Ω = 134.9 Ω
The rms value of the current in the circuit,
`"i"_"rms" = "V"_"rms"/"Z" = 110/134.9` A = 0.8154 A
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