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25 May, 20:35

The capacitor can withstand a peak voltage of 600 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude VmaxVmaxV_max can the source have if the maximum capacitor voltage is not exceeded?

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Answers (2)
  1. 25 May, 22:22
    0
    The question is not complete and the first part of the question says;

    In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 henrys, and the capacitance is 1.20*10^ (-2) microfarads.

    Answer:

    Vs = 42.65 V

    Explanation:

    Formula for resonance is given as;

    f = 1 / (2π (√LC))

    Where L is inductance in henrys

    While C is Capacitance in farads

    In this question;

    C = 1.20*10^ (-2) microfarads = 1.2 x 10^ (-8) Farads

    L = 0.38 H

    Thus,

    f = 1 / (2π (√1.20*10^ (-8) x 0.38))

    f = 1/0.00042428951

    f = 2356.88 Hz

    Now, capacitive resistance Xc is given as;

    Xc = 1 / (2πfC)

    Xc = 1 / (2π x 2356.88 x 1.2 x 10^ (-8)) = 5627.32 ohms

    Since the capacitors can withstand a peak voltage of 600V.

    Thus Vc = IXc

    Where I is current thus, Vc/Xc = I

    Also, for the series, Vs/R = I

    Thus, Vc/Xc = Vs/R

    So, Vc = 600 V, Xc = 5627.32 ohms while R = 400 ohms

    Thus,

    Making Vs the subject,

    (Vc/Xc) x R = Vs

    Thus,

    (600/5627.32) x 400 = Vs

    Vs = 42.65 V
  2. 25 May, 22:55
    0
    426.5 V

    Explanation:

    The complete question is:

    In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 henrys, and the capacitance is 1.20*10-2 microfarads.

    The capacitor can withstand a peak voltage of 600 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude V_max can the source have if the maximum capacitor voltage is not exceeded?

    at resonance

    XL = Xc

    2πfL=1/2πfC

    2*π*f*0.38 = 1 / (2π*f*0.0000012)

    f = 235.8 Hz

    XL = 2π*235.8*0.38 = 562.7

    Xc = 1 / (2π*235.8*0.0000012) = 562.7

    at resonance

    Z = 400

    Vc = - j562.7 / 400 * Vs

    600 = - j562.7/400 * Vs

    Vs = 426.5V
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