Question

An ac source is rated at \(220 \mathrm{~V}, 50 \mathrm{~Hz}\). The time taken for voltage to change from its peak value to zero is, (a) \(50 \mathrm{sec}\) (b) \(0.02 \mathrm{sec}\) (c) \(5 \mathrm{sec}\) (d) \(5 \times 10^{-3} \mathrm{sec}\)

Short answer

The time taken for the voltage to change from its peak value to zero is found by first calculating the time period of one cycle using the formula \(T = \frac{1}{f}\), where f is the given frequency (50 Hz). The time period is then divided by 4, as the voltage decreases to zero within a quarter of a cycle. Thus, the final answer is \(t = \frac{1}{4} \times \frac{1}{50} = 5 \times 10^{-3} \mathrm{~sec}\), which corresponds to option (d).

Step-by-step solution

Calculate the time period of one cycle

The frequency (f) of an AC voltage source is the number of complete cycles it goes through in one second. The relationship between frequency and time period (T) is: \(T = \frac{1}{f}\) Given the frequency (f) is 50 Hz, we can find the time period with the formula above: \(T = \frac{1}{50}\)

Determine the time taken for the voltage to change from its peak value to zero

As previously mentioned, the voltage will decrease to zero within a quarter of a cycle. Therefore, we need to calculate 1/4 of the time period: \(t = \frac{1}{4}T\) Now insert the calculated time period (T) we found in step 1: \(t = \frac{1}{4} \times \frac{1}{50}\)

Calculate the final result and select the correct option

Now, we only need to do the calculation: \(t = \frac{1}{4} \times \frac{1}{50} = \frac{1}{200} = 5 \times 10^{-3} \mathrm{~sec}\) So the correct answer is (d) 5 × 10⁻³ sec.

Key concepts

Time Period Calculation

The first step in analyzing an AC circuit is to understand the time period of the AC waveform. The time period (\(T\)) is the duration of one complete cycle of the waveform. This is essential knowledge because many AC functions depend on this periodicity. The time period is inversely related to the frequency (\(f\)). You can calculate the time period using the equation:

• \(T = \frac{1}{f}\)

Given that our AC source has a frequency of 50 Hz, this means it completes 50 cycles every second. So, by placing this into our formula, we calculate:\(T = \frac{1}{50} = 0.02\) seconds.
Thus, every cycle of this AC waveform lasts 0.02 seconds, which helps us in further analysis of its behavior.

Frequency and Voltage Relationship

The frequency of an AC source not only tells us how fast it cycles but also affects the voltage characteristics over time. In our scenario, we have an AC source rated at 220 V and 50 Hz. Here's how frequency impacts voltage behavior:

• The frequency defines how many times the voltage reaches its peak in a second.
• At 50 Hz, the voltage reaches its maximum value of 220 V and then decreases to zero, both multiple times each second.

The relationship between frequency and voltage is crucial in understanding the timing of voltage changes in AC, and this timing is predictable based on the frequency. Hence, changing the frequency will affect how quickly the voltage changes from its peak to zero.

Time for Voltage Change in AC

When considering the time taken for an AC voltage to change from its peak value to zero, it's important to know that this voltage change occurs over a quarter of the full cycle. In AC waveforms, like sine waves, the voltage goes from peak to zero, back to peak in a smooth, regular pattern due to its periodic nature.Since we know that the time period for one complete cycle is 0.02 seconds:

• We calculate the time to shift from peak to zero as a quarter of this period.
• Using the calculation: \(t = \frac{1}{4} \times T = \frac{1}{4} \times 0.02 \) seconds, we find \(t = 0.005\) seconds or \(5 \times 10^{-3}\) seconds.

Therefore, each time the AC waveform goes from peak voltage to zero, it does so in 0.005 seconds. This periodical nature allows easy prediction of voltage behavior over time, which is critical in AC circuit analysis.