You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source

Question

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 220 rad/s .
Part A: What is the impedance of the circuit? ( Answer: Z = ? Ω )
Part B: What is the current amplitude? ( Answer: I = ? A )
Part C: What is the voltage amplitude across the resistor? ( Answer: VR = ? V )
Part D: What is the voltage amplitudes across the inductor? ( Answer: VL = ? V )
Part E: What is the phase angle ϕ of the source voltage with respect to the current? ( Answer: ϕ = ? degrees )
Part F: Does the source voltage lag or lead the current? ( Answer: the voltage lags the current OR the voltage leads the current )

in progress 0
Yến Oanh 4 years 2021-08-18T22:04:08+00:00 1 Answers 21 views 0

Answers ( )

  1. ngockhue
    0
    2021-08-18T22:06:05+00:00
    Reply

    Answer:

    A.  Z = 185.87Ω

    B.  I  =  0.16A

    C.  V = 1mV

    D.  VL = 68.8V

    E.  Ф = 30.59°

    Explanation:

    A. The impedance of a RL circuit is given by the following formula:

    Z=\sqrt{R^2+\omega^2L^2}       (1)

    R: resistance of the circuit = 160-Ω

    w: angular frequency = 220 rad/s

    L: inductance of the circuit = 0.430H

    You replace in the equation (1):

    Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega

    The impedance of the circuit is 185.87Ω

    B. The current amplitude is:

    I=\frac{V}{Z}                     (2)

    V: voltage amplitude = 30.0V

    I=\frac{30.0V}{185.87\Omega}=0.16A

    The current amplitude is 0.16A

    C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

    V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV            (3)

    D. The voltage across the inductor is:

    V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V

    E. The phase difference is given by:

    \phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )