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Transcribed Image Text:**Problem 11: Alternating Voltage Analysis**
An alternating voltage, \( v \), has a periodic time of \( 0.01 \, \text{s} \) and a peak value of \( 40 \, \text{V} \). When time \( t \) is zero, \( v = -20 \, \text{V} \). Express the instantaneous voltage in the form \( v = V_m \sin(\omega t \pm \phi) \).
**Explanation:**
- **Periodic Time**: The periodic time of the voltage is \( 0.01 \, \text{s} \). This means the voltage wave completes one full cycle every \( 0.01 \, \text{s} \).
- **Peak Value**: The maximum (or amplitude) of the voltage is \( 40 \, \text{V} \).
- **Initial Condition**: At time \( t = 0 \), the voltage \( v \) is \( -20 \, \text{V} \).
- **Objective**: To express the instantaneous voltage using the given sinusoidal function format.
This problem is an exercise in expressing an AC voltage in its sinusoidal form, which is commonly used in electrical engineering to describe alternating current and voltage waves. The form \( v = V_m \sin(\omega t \pm \phi) \) represents the amplitude \( V_m \), angular frequency \( \omega \), time \( t \), and phase shift \( \phi \).
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