A 13.6- resistor, an 11.9-μF capacitor, and a 19.1-mH inductor are connected in series with a 117-V generator. (a) At what frequency

Question

A 13.6- resistor, an 11.9-μF capacitor, and a 19.1-mH inductor are connected in series with a 117-V generator.
(a) At what frequency is the current a maximum?
(b) What is the maximum value of the RMS current?
Note: The ac current and voltage are RMS values and power is an average value unless indicated otherwise.

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Huyền Thanh 4 years 2021-07-31T06:15:57+00:00 2 Answers 11 views 0

Answers ( )

  1. lethu
    0
    2021-07-31T06:17:43+00:00
    Reply

    Answer:

    a) Current is maximum at frequency, f₀ = 333.83 Hz

    b) Maximum current = 12.17 A

    Explanation:

    Inductance, L = 19.1 mH = 19.1 * 10⁻³ H

    Capacitance, C = 11.9 μF =11.9 * 10⁻⁶ F

    a) Current is maximum at resonant frequency, f₀

    f_{0} = \frac{1}{2\pi\sqrt{LC} }

    f_{0} = \frac{1}{2\pi\sqrt{11.9 * 10^{-6}* 19.1 * 10^{-3} } }

    f_{0} = 333.83 Hz

    b) Maximum value of the RMS current

    V_{RMS} = 117 V\\V_{max} = \sqrt{2} V_{RMS}\\V_{max} = \sqrt{2} * 117\\V_{max} = 165.46 V

    I_{max} = \frac{V_{max} }{R} \\I_{max} = \frac{165.46}{13.6} \\I_{max} = 12.17 A

  2. thienthanh
    0
    2021-07-31T06:17:44+00:00
    Reply

    Answer:

    Explanation:

    Given the following information,

    Resistor of resistance R = 13.6Ω

    Capacitor of capacitance C = 11.9-μF

    C = 11.9 × 10^ -6 F

    Inductor of inductance L = 19.1-mH

    L = 19.1 ×10^-3 H

    All this connected in series to a generator that generates Vrms= 117V

    Vo = Vrms√2 = 117√2

    Vo = 165.463V

    a. Frequency for maximum current?

    Maximum current occurs at resonance

    I.e Xc = XL

    At maximum current, the frequency is given as

    f = 1/(2π√LC)

    Then,

    f = 1/(2π√(19.1×10^-3 × 11.9×10^-6)

    f = 1/(2π√(2.2729×10^-7))

    f = 1/(2π × 4.77 ×10^-4)

    f = 333.83Hz

    Then, the frequency is 333.83Hz.

    b. Since we know the frequency,

    Then, we need to find the capacitive and inductive reactance

    Capacitive reactance

    Xc = 1/2πfC

    Xc = 1/(2π × 338.83 × 11.9×10^-6)

    Xc = 1/ 0.024961

    Xc = 40.1Ω

    Also, Inductive reactance

    XL = 2πfL

    XL = 2π × 333.83 × 19.1×10^-3

    XL = 40.1Ω

    As expected, Xc=XL, resonance

    Then, the impedance in AC circuit is given as

    Z = √ (R² + (Xc—XL)²)

    Z = √ 13.6² + (40.1-40.1)²)

    Z = √13.6²

    Z = 13.6 ohms

    Then, using ohms las

    V = IZ

    Then, I = Vo/Z

    Io = 165.46/13.6

    Io = 12.17Amps

    The current is 12.17 A

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