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.