Question

Assuming a 60hz system, how much time in seconds is the peak current behind

Answer 1

In a 60Hz AC system, the peak current occurs 0.00417 seconds after the zero-crossing point of the waveform, and for a circuit with a 15.0 A RMS current, the peak current would be approximately 21.2 A.

Step-by-step explanation:

The question relates to alternating current (AC) systems functioning at 60Hz.

In such a system, if we want to find the time that the peak current lags or leads, we have to understand the relationship between root mean square (RMS) and peak values in AC circuits.

Since we are speaking about a 60Hz system, a complete cycle takes 1/60 seconds.

The peak current or voltage occurs at the quarter cycle, which is 1/240 seconds or approximately 0.00417 seconds behind for current reaching its peak value after the zero-crossing point.

If the circuit breaker trips at an RMS current of 15.0 A, then the corresponding peak current would be higher by a factor of the square root of two (√2), as the RMS value represents the value of an equivalent direct current (DC) that would deliver the same power.

Simply multiply the RMS value by √2 to determine the peak current.

This calculation yields a peak current of approximately 21.2 A for a 15 A RMS value.

Answer 2

Accelerating uniformly from rest along an on-ramp to merge into interstate traffic at 60 mi/h on two occasions implies two different acceleration durations and magnitudes, as the distance traveled will likely be different. To precisely determine the acceleration values and distances, additional information such as the time taken for acceleration or specific acceleration profiles is needed.

Step-by-step explanation:

When accelerating uniformly from rest, the kinematic equation
`\(v = u + at\)` can be used, where (v) is the final velocity, (u) is the initial velocity (which is 0 when starting from rest), (a) is the acceleration, and (t) is the time of acceleration. If you're accelerating to a final speed of 60 mi/h, it's important to convert this to consistent units, such as feet per second. One mile is equivalent to 5280 feet, and one hour is equivalent to 3600 seconds. Therefore, 60 mi/h is approximately 88 feet per second.

Without specific information about the duration of acceleration or the specific acceleration profile, it's challenging to provide precise acceleration values or distances. However, the general approach involves using the kinematic equations to find either the acceleration magnitude or the distance traveled during acceleration. Additional details about the scenario, such as the time taken to reach the desired speed or the acceleration rate, would allow for a more detailed analysis.

This type of problem is common in physics and engineering, where understanding motion and acceleration is essential. It demonstrates the application of kinematic equations to real-world scenarios, such as merging into traffic, where precise control of acceleration is important for safety and efficiency.