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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, impedanc - Physics

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प्रश्न

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?

संख्यात्मक

उत्तर

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 = `|omega"L" - 1/(omega"C")| = |2pi"fL" - 1/(2pi"fC")|`

`= |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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अध्याय 13: AC Circuits - Exercises [पृष्ठ ३०५]

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बालभारती Physics [English] 12 Standard HSC Maharashtra State Board
अध्याय 13 AC Circuits
Exercises | Q 17 | पृष्ठ ३०५

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