Given v (t) = 4 Sin 200 πt. Sketch the voltage function of time. Determine the waveform as a rms value of the wave form and its period. Y
Given v (t) = 4 Sin 200 πt. Sketch the voltage function of time. Determine the waveform as a rms value of the wave form and its period. Y
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![**Problem Statement:**
Given \( v(t) = 4 \sin(200\pi t) \).
1. **Sketch the voltage waveform as a function of time.**
2. **Determine the RMS value of the waveform and its period.**
**Solution:**
1. **Sketching the Voltage Waveform:**
For \( v(t) = 4 \sin(200\pi t) \), the waveform is a sinusoidal function. The amplitude is 4, which means the peak values of the sine wave will be +4 and -4. The frequency \( f \) can be found from the term \( 200\pi t \), where \( \omega = 2\pi f \). Here, \( \omega = 200\pi \), so:
\[
f = \frac{200\pi}{2\pi} = 100 \, \text{Hz}
\]
2. **Determine the RMS Value and Period:**
- **RMS Value:**
The RMS (Root Mean Square) value for a sine wave \( v(t) = A \sin(\omega t) \) is given by:
\[
V_{\text{RMS}} = \frac{A}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.83
\]
- **Period:**
The period \( T \) is the reciprocal of the frequency \( f \):
\[
T = \frac{1}{f} = \frac{1}{100} = 0.01 \, \text{seconds} \, (10 \, \text{ms})
\]
The waveform is therefore a 100 Hz sine wave with a period of 10 ms and an RMS value of approximately 2.83 volts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a1a8c69-ac20-486b-9f67-8f66504c5494%2F44bac51a-8b9f-4aee-afdb-63c1b47fa374%2F8g0ld8mp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given \( v(t) = 4 \sin(200\pi t) \).
1. **Sketch the voltage waveform as a function of time.**
2. **Determine the RMS value of the waveform and its period.**
**Solution:**
1. **Sketching the Voltage Waveform:**
For \( v(t) = 4 \sin(200\pi t) \), the waveform is a sinusoidal function. The amplitude is 4, which means the peak values of the sine wave will be +4 and -4. The frequency \( f \) can be found from the term \( 200\pi t \), where \( \omega = 2\pi f \). Here, \( \omega = 200\pi \), so:
\[
f = \frac{200\pi}{2\pi} = 100 \, \text{Hz}
\]
2. **Determine the RMS Value and Period:**
- **RMS Value:**
The RMS (Root Mean Square) value for a sine wave \( v(t) = A \sin(\omega t) \) is given by:
\[
V_{\text{RMS}} = \frac{A}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.83
\]
- **Period:**
The period \( T \) is the reciprocal of the frequency \( f \):
\[
T = \frac{1}{f} = \frac{1}{100} = 0.01 \, \text{seconds} \, (10 \, \text{ms})
\]
The waveform is therefore a 100 Hz sine wave with a period of 10 ms and an RMS value of approximately 2.83 volts.
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