Question:
The time taken by AC of 50 Hz in reaching from zero to the maximum value is:
The time taken by AC of 50 Hz in reaching from zero to the maximum value is:
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The time taken for the AC voltage to reach its maximum value is one-fourth of the time period of the AC cycle.
Updated On: Apr 22, 2026
- \( 50 \times 10^{-3} \, \text{s} \)
- \( 5 \times 10^{-3} \, \text{s} \)
- \( 1 \times 10^{-3} \, \text{s} \)
- \( 2 \times 10^{-3} \, \text{s} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the formula for AC time.
In an AC circuit, the time taken for the voltage to reach its maximum value is a fraction of the period of the wave. The time period \( T \) for an AC with frequency \( f \) is given by: \[ T = \frac{1}{f} \]
Step 2: Apply the given frequency.
We are given the frequency \( f = 50 \, \text{Hz} \), so the time period is: \[ T = \frac{1}{50} = 0.02 \, \text{s} \]
Step 3: Calculate the time to reach maximum value.
The time taken for the voltage to reach its maximum value is a quarter of the time period: \[ \text{Time to maximum} = \frac{T}{4} = \frac{0.02}{4} = 5 \times 10^{-3} \, \text{s} \]
In an AC circuit, the time taken for the voltage to reach its maximum value is a fraction of the period of the wave. The time period \( T \) for an AC with frequency \( f \) is given by: \[ T = \frac{1}{f} \]
Step 2: Apply the given frequency.
We are given the frequency \( f = 50 \, \text{Hz} \), so the time period is: \[ T = \frac{1}{50} = 0.02 \, \text{s} \]
Step 3: Calculate the time to reach maximum value.
The time taken for the voltage to reach its maximum value is a quarter of the time period: \[ \text{Time to maximum} = \frac{T}{4} = \frac{0.02}{4} = 5 \times 10^{-3} \, \text{s} \]
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