You have a 192 −Ω−Ω resistor, a 0.409 −H−H inductor, a 4.95 −μF−μF capacitor, and a variable-frequency ac source with an amplitude of 3.02 V

Question

You have a 192 −Ω−Ω resistor, a 0.409 −H−H inductor, a 4.95 −μF−μF capacitor, and a variable-frequency ac source with an amplitude of 3.02 VV . You connect all four elements together to form a series circuit.

(a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency?
(b) What will be the current amplitude at an angular frequency of 400 rad/s? At this frequency, will the source voltage lead or lag the current?

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Edana Edana 2 months 2021-07-24T01:40:25+00:00 1 Answers 1 views 0

Answers ( )

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    2021-07-24T01:41:46+00:00

    Answer:

    Explanation:

    Given that,

    The resistance of the resistor is

    R = 192 Ω

    The inductance of the inductor is

    L = 0.409 H

    The capacitance of the capacitor is

    C = 4.95 μF

    a. At what frequency is current max?

    The current will be maximum at resonance, and resonance frequency is given as

    f = 1/2π√L•C

    f = 1/2π√(0.409×4.95×10^-6)

    f = 111.86Hz

    b. Currency at frequency 400rad/s.

    w = 400rad/s

    So, we need to find the inductive reactance

    XL = wL

    XL = 400 ×0.409

    XL = 163.6 ohms

    We also need the capacitive reacttance

    XC = 1/wC

    XC = 1/400×4.95×10^-6

    XC = 505.05 ohms

    Then, the impedance of the circuit is given as

    Z = √(R²+(XL-XC)²)

    Z = √(192²+(505.05—163.6)²)

    Z = √(192²+341.9²)

    Z = 392.12 ohms

    Then, the current that flows can be calculated using

    V= IZ

    I = V/Z

    I = 3.02/392.12

    I = 7.7 × 10^-3 A

    I = 7.7 mA

    Since Xc>XL, then, the source voltage lags the current

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