The voltage signal from a standard european wall socket can be described by the equation V(t) = 325Sin100(pie)t, where t is time in seconds and v is voltage at time t

A) find the maximum and minimum voltage levels and times at which they occur

B) determine
i) the petiod, T, in seconds
ii) the frequency, f, in Hertz
iii) the amplitude, A, in volts

To find the maximum and minimum voltage levels and times at which they occur, we need to analyze the equation V(t) = 325Sin(100πt).

A) Maximum and Minimum Voltage Levels and Times:
The general form of a sinusoidal function is A * Sin(Bx + C), where A represents the amplitude, B represents the frequency, and C represents a phase shift.

1. Amplitude (A):
The amplitude of a sinusoidal function is the maximum displacement from the mean or average value. In this case, the amplitude is 325 volts.

2. Maximum Voltage:
The maximum voltage occurs when Sin(Bx + C) equals +1. In our equation, Sin(100πt) reaches its maximum value at t = 0. Therefore, the maximum voltage level is 325 volts at time t = 0.

3. Minimum Voltage:
The minimum voltage occurs when Sin(Bx + C) equals -1. In our equation, Sin(100πt) reaches its minimum value when Sin(100πt) = -1. This occurs when 100πt = -π/2 or t = -1/200 seconds (or t = -0.005 seconds). Hence, the minimum voltage level is -325 volts at time t = -0.005 seconds.

B) Period (T), Frequency (f), and Amplitude (A):
1. Period (T):
The period of a sinusoidal function is the time it takes for the waveform to complete one full cycle. In our equation, the argument of sine is 100πt. To determine the period, we can set 100πt equal to 2π (one full cycle): 100πt = 2π. Solving for t, we get t = 1/50 seconds. Therefore, the period (T) is 1/50 seconds.

2. Frequency (f):
The frequency of a sinusoidal function is the number of cycles it completes in one second. Since the period is the reciprocal of the frequency, f = 1/T. Substituting the value of T we found earlier, the frequency (f) is 50 Hz.

3. Amplitude (A):
We have already determined the amplitude to be 325 volts.

To summarize:
i) The period (T) is 1/50 seconds.
ii) The frequency (f) is 50 Hz.
iii) The amplitude (A) is 325 volts.

A) To find the maximum and minimum voltage levels and the times at which they occur, we can look at the equation V(t) = 325sin(100πt).

1. Maximum voltage level: The maximum voltage level occurs when the sine function reaches its maximum value of 1. Therefore, the maximum voltage level is 325 volts.

2. Minimum voltage level: The minimum voltage level occurs when the sine function reaches its minimum value of -1. Therefore, the minimum voltage level is -325 volts.

To find the times at which these extreme voltage levels occur, we can use the fact that the sine function repeats itself every 2π radians or 360 degrees. In other words, the sine function has a period of T = 2π/ω, where ω is the angular frequency.

B) Now, let's calculate the period, frequency, and amplitude.

i) Period, T: We can determine the period by recognizing that the coefficient of t in the sine function (100π) corresponds to the angular frequency ω. The period T can be found as T = 2π/ω. Plugging in the value, we have:

T = 2π/(100π) = 1/50 seconds

Therefore, the period T is 1/50 seconds.

ii) Frequency, f: The frequency is the reciprocal of the period, which means f = 1/T. Plugging in the value we found for T, we have:

f = 1/(1/50) = 50 Hz

So, the frequency f is 50 Hz.

iii) Amplitude, A: The amplitude of a sine function is the maximum value it reaches. In this case, the maximum voltage level of 325 volts is the amplitude.

Therefore, the amplitude A is 325 volts.