A 40 kHz sinusoidal voltage has zero phase angle and a maximum amplitude of 25 mV. When this voltage is applied across the terminals of a capacitor, the resulting steady-state current has a maximum amplitude of 628.32 μA. Part A What is the frequency of the current in radians per second? Express your answer in radians per second to three significant figures. Submit Part B 0₁ = VAΣ vec Submit VI What is the phase angle of the current? Express your answer in degrees to three significant figures. Request Answer VAΣ vec Request Answer ? ? rad/s

A 40 kHz sinusoidal voltage has zero phase angle and a maximum amplitude of 25 mV. When this voltage is applied across the terminals of a capacitor, the resulting steady-state current has a maximum amplitude of 628.32 μA. Part A What is the frequency of the current in radians per second? Express your answer in radians per second to three significant figures. Submit Part B 0₁ = VAΣ vec Submit VI What is the phase angle of the current? Express your answer in degrees to three significant figures. Request Answer VAΣ vec Request Answer ? ? rad/s

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.27P: An industrial load consisting of a bank of induction motors consumes 50 kW at a power factor of 0.8...

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Read A 70-Vac source has the following waveform. Determine:a. Vpkb. Vpk-pkc. Vrmsd. Periode. Frequencyf. Angular Velocityg. Phase Angleh. equation of the waveform (in time domain)i. the instantaneous voltage when t = 120 msj. the angle (1st occurrence) after t = 0 when the voltage is +80 Vk. the time (2nd occurrence) after t = 0 when the voltage is –10 VAnswers:a. 98.9949 Vb. 197.9899 Vc. 70 Vd. 100 mse. 10 Hzf. 20π rad/secg. 125.0876°h. v(t) = 98.9949sin(20πt + 125.0876°)i. -29.0880 Vj. 0.99959°k. 63.643 ms Give the solution for J and K ONLY ans (0.99959 degrees and 63.643 ms respectively)
Give the solution for J and K ONLY ans (0.99959 degrees and 63.643 ms respectively).Read A 70-Vac source has the following waveform. Determine:a. Vpkb. Vpk-pkc. Vrmsd. Periode. Frequencyf. Angular Velocityg. Phase Angleh. equation of the waveform (in time domain)i. the instantaneous voltage when t = 120 msj. the angle (1st occurrence) after t = 0 when the voltage is +80 Vk. the time (2nd occurrence) after t = 0 when the voltage is –10 VAnswers:a. 98.9949 Vb. 197.9899 Vc. 70 Vd. 100 mse. 10 Hzf. 20π rad/secg. 125.0876°h. v(t) = 98.9949sin(20πt + 125.0876°)i. -29.0880 Vj. 0.99959°k. 63.643 ms
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