Question

The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value? ____.

• A. 1/60 s
• B. 1/240 s
• C. 1/30 s
• D. 1/120 s

Hint:

A full cycle is $360^\circ$ ($T$). Peak happens at $90^\circ$, which is exactly $1/4$ of the cycle. Therefore, time to peak is always $1/(4f)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
This problem belongs to "Alternating Current" (AC). It involves understanding the time-domain behavior of a sinusoidal signal. An AC signal oscillates between positive and negative peaks, and its "frequency" tells us how many full cycles occur every second. The "Time Period" is the duration of one such full cycle.
Step 2: Key Formulas and Approach:

Time Period ($T$) = $1 / \text{frequency } (f)$.
A full cycle ($360^\circ$ or $2\pi$ radians) takes time $T$.
In a sine wave starting at zero ($I = I_0 \sin \omega t$), the first peak occurs at $90^\circ$ (or $\pi/2$ radians).
Therefore, time to peak ($t_{peak}$) = $T / 4$.

Step 3: Detailed Explanation:

Identify given values: Peak current $I_0 = 5 \text{ A}$ (not needed for the calculation) and frequency $f = 60 \text{ Hz}$.
Calculate the Time Period: \[ T = \frac{1}{f} = \frac{1}{60} \text{ seconds} \]
Determine the time to reach the first peak: A full AC cycle consists of:
0 to Peak (1/4 cycle)
Peak back to 0 (1/4 cycle)
0 to Negative Peak (1/4 cycle)
Negative Peak back to 0 (1/4 cycle)

Thus, reaching the peak starting from zero takes one quarter of the total period: \[ t = \frac{T}{4} = \frac{1/60}{4} \] \[ t = \frac{1}{60 \times 4} = \frac{1}{240} \text{ seconds} \]
Step 4: Final Answer:
The current takes 1/240 s to reach the peak value.