23.03.2023 - 17:36

You have a 170-ohm resistor and a 0.370 H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 33.0 V and an angular fre

Question:

You have a 170-ohm resistor and a 0.370 H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 33.0 V and an angular frequency of 270 rad/s.

(a) What is the impedance of the circuit?
(b) What is the current amplitude?
(c) What is the voltage amplitude across the resistor?
(d) What is the voltage amplitudes across the inductor?
(e) What is the phase angle {eq}phi {/eq} of the source voltage with respect to the current?
(f) Construct the phasor diagram.

Answers (1)
  • garik20101
    April 10, 2023 в 23:53

    (a) The impedance of the circuit is the total opposition to the flow of current and can be calculated using the formula:

    Z = sqrt(R^2 + X_L^2)

    where R is the resistance, X_L is the inductive reactance, given by:

    X_L = 2πfL

    where f is the frequency and L is the inductance.

    Plugging in the values, we get:

    X_L = 2π(270 rad/s)(0.370 H) = 229.68 ohms

    Z = sqrt((170 ohms)^2 + (229.68 ohms)^2) = 286.66 ohms

    Therefore, the impedance of the circuit is 286.66 ohms.

    (b) The current amplitude can be calculated using Ohm's Law, which states that the current is equal to the voltage divided by the impedance:

    I = V/Z

    Plugging in the values, we get:

    I = (33.0 V) / (286.66 ohms) = 0.115 A

    Therefore, the current amplitude is 0.115 A.

    (c) The voltage amplitude across the resistor can be calculated using Ohm's Law:

    V_R = I R

    Plugging in the values, we get:

    V_R = (0.115 A) (170 ohms) = 19.55 V

    Therefore, the voltage amplitude across the resistor is 19.55 V.

    (d) The voltage amplitude across the inductor can be calculated using Ohm's Law:

    V_L = I X_L

    Plugging in the values, we get:

    V_L = (0.115 A) (229.68 ohms) = 26.40 V

    Therefore, the voltage amplitude across the inductor is 26.40 V.

    (e) The phase angle {eq}phi{/eq} of the source voltage with respect to the current can be calculated using the formula:

    tan({eq}phi{/eq}) = X_L / R

    Plugging in the values, we get:

    tan({eq}phi{/eq}) = 229.68 ohms / 170 ohms = 1.35

    Using a calculator, we find that:

    {eq}phi{/eq} = 52.58 degrees

    Therefore, the phase angle {eq}phi{/eq} of the source voltage with respect to the current is 52.58 degrees.

    (f) The phasor diagram can be constructed by drawing a right triangle where the hypotenuse represents the impedance Z, the adjacent side represents the resistance R, and the opposite side represents the inductive reactance X_L. The angle between the hypotenuse and the adjacent side represents the phase angle {eq}phi{/eq}. The phasor diagram for this circuit is shown below:

    bash
    | Z | X_L | | | | | | | {eq}phi{/eq} -----|----------------- | | R | | | | | |_______________ Voltage

    The hypotenuse of the triangle is the impedance Z = 286.66 ohms, the adjacent side is the resistance R = 170 ohms, and the opposite side is the inductive reactance X_L = 229.68 ohms. The phase angle {eq}phi{/eq} is 52.58 degrees, which is the angle between the hypotenuse and the adjacent side. The voltage amplitude is shown at the end of the hypotenuse.

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