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P5.6. A sinusoidal voltage has a peak value of 15 V, has a frequency of 125 Hz, and crosses zero with positive slope at t =1 ms. Write an expression for the voltage.

P5.6. A sinusoidal voltage has a peak value of 15 V, has a frequency of 125 Hz, and crosses zero with positive slope at t =1 ms. Write an expression for the voltage.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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**P5.6** A sinusoidal voltage has a peak value of 15 V, has a frequency of 125 Hz, and crosses zero with positive slope at \( t = 1 \, \text{ms} \). Write an expression for the voltage. --- In this problem, we are given a sinusoidal voltage signal with specific characteristics: - **Peak Value**: 15 Volts - **Frequency**: 125 Hertz - **Zero Crossing**: Occurs with a positive slope at \( t = 1 \, \text{ms} \) To find the expression for the voltage, we will use the general form of a sinusoidal function: \[ v(t) = V_m \sin(2\pi ft + \phi) \] Where: - \( v(t) \) is the instantaneous voltage at time \( t \). - \( V_m \) is the peak voltage (15 V in this case). - \( f \) is the frequency (125 Hz). - \( \phi \) is the phase angle, which needs to be determined. Since the voltage is zero and crossing with a positive slope at \( t = 1 \, \text{ms} \), we need to adjust the phase angle \( \phi \) so that the sine function satisfies these conditions.
Transcribed Image Text:**P5.6** A sinusoidal voltage has a peak value of 15 V, has a frequency of 125 Hz, and crosses zero with positive slope at \( t = 1 \, \text{ms} \). Write an expression for the voltage. --- In this problem, we are given a sinusoidal voltage signal with specific characteristics: - **Peak Value**: 15 Volts - **Frequency**: 125 Hertz - **Zero Crossing**: Occurs with a positive slope at \( t = 1 \, \text{ms} \) To find the expression for the voltage, we will use the general form of a sinusoidal function: \[ v(t) = V_m \sin(2\pi ft + \phi) \] Where: - \( v(t) \) is the instantaneous voltage at time \( t \). - \( V_m \) is the peak voltage (15 V in this case). - \( f \) is the frequency (125 Hz). - \( \phi \) is the phase angle, which needs to be determined. Since the voltage is zero and crossing with a positive slope at \( t = 1 \, \text{ms} \), we need to adjust the phase angle \( \phi \) so that the sine function satisfies these conditions.
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