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Question
A 25 μF capacitor, a 0.10 H inductor, and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current, and phase angle of the circuit?
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Solution
Data: C = 25 µF = 25 x 10-6 F
L = 0.10 H
R = 25Ω
e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with
e = e0 sin (2πft)
we get, the frequency of the alternating emf as
f = `314/(2pi) = 314/(2(3.14))` = 50 Hz
Reactance = `|omegaL - 1/(omegaC)| = |2pifL - 1/(2pifC)|`
`= |2(3.14)(50)(0.10) - 1/(2(3.14)(50)(25 xx 10^-6))|`
`= |31.4 - (100 xx 10^2)/((3.14)(25))|` = |31.4 - 127.4|
= 96 Ω
`Z^2 = R^2 + (omega"L" - 1/(omega"C"))^2 = (25)^2 + (96)^2 = 9841 Omega^2`
∴ Impedance, Z = `sqrt9841`Ω = 99.2 Ω
Peak current, `i_0 = e_0/Z = 310/99.2`A
∴ `i_"rms" = "i"_0/sqrt2 = 310/((1.414)(99.2))`A = 2.21 A
cos Φ = `R/Z = 25/99.2 = 0.2520`
∴ Phase angle,
Φ = cos-1(0.2520) = 75.40° = 1.316 rad
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