The voltage signal from a standard European wall socket can be described by the equation V(t)=325sin100 pi x ,

where t is time, in seconds and V is the voltage at time, t.

Find the max and min voltage levels and the times at which they occur.

Thanks in advance!

To find the maximum and minimum voltage levels, we can look at the amplitude of the sine function.

The amplitude of a sinusoidal function is equal to the maximum displacement from the equilibrium position. For the given equation V(t) = 325sin(100πt), the amplitude is 325.

So, the maximum voltage level is 325 volts and the minimum voltage level is -325 volts.

To find the times at which these voltage levels occur, we need to determine the values of t that satisfy the equation V(t) = ±325.

For the maximum voltage level, V(t) = 325:
325 = 325sin(100πt)

To solve this equation, we can divide both sides by 325 and take the inverse sine (or arcsine) of both sides:
sin^(-1)(1) = 100πt

Since the arcsine of 1 is π/2, we have:
π/2 = 100πt

Simplifying, we get:
t = (π/2) / (100π)
t = 1 / (200)

Therefore, the maximum voltage level occurs at t = 1 / (200) seconds.

For the minimum voltage level, V(t) = -325:
-325 = 325sin(100πt)

Again, dividing both sides by 325 and taking the inverse sine, we get:
sin^(-1)(-1) = 100πt

Since the arcsine of -1 is -π/2, we have:
-π/2 = 100πt

Simplifying, we get:
t = (-π/2) / (100π)
t = -1 / (200)

Therefore, the minimum voltage level occurs at t = -1 / (200) seconds.

To summarize, the maximum voltage level of 325 volts occurs at t = 1 / (200) seconds, and the minimum voltage level of -325 volts occurs at t = -1 / (200) seconds.

To find the max and min voltage levels, we need to determine the amplitude of the sine function. The amplitude is the maximum value reached by the function, which in this case represents the maximum voltage level.

In the equation V(t) = 325sin(100πt), the coefficient of the sine function, 325, represents the amplitude. Therefore, the maximum voltage level is 325 volts.

To find the minimum voltage level, we need to consider that the sine function oscillates between positive and negative values. The minimum voltage level will be the negative of the maximum voltage level, which is -325 volts.

Now, to determine the times at which these maximum and minimum voltage levels occur, we look at the argument of the sine function, 100πt. The sine function oscillates with a period of 2π. Thus, the time it takes to complete one cycle is T = 2π/100π = 1/50 seconds.

For the maximum voltage level, it occurs when the argument of the sine function is equal to π/2 or any odd multiple of π/2. So, we can write:

100πt = π/2 + kπ, where k is an odd integer.

Simplifying, we have:
t = (π/2 + kπ)/(100π) = (1 + 100k)/200.

For the minimum voltage level (negative maximum), it occurs when the argument of the sine function is equal to 3π/2 or any odd multiple of 3π/2. So, we can write:

100πt = 3π/2 + kπ, where k is an odd integer.

Simplifying, we have:
t = (3π/2 + kπ)/(100π) = (3 + 100k)/200.

Therefore, to find the times at which the maximum and minimum voltage levels occur, substitute odd values for k. For example, when k = 1, we get the first maximum and minimum voltages. Plugging it into the formulas, we find:

For the first maximum voltage, t = (1 + 100 * 1)/200 = 101/200 seconds.
For the first minimum voltage, t = (3 + 100 * 1)/200 = 103/200 seconds.

You can change the value of k to find the times for subsequent maximum and minimum voltage levels.

clearly the max and mins are 325 and -325 just knowing what the graph looks like.

The period of sin(100πx) is 2π/(100π) = 1/50 sec
knowing that the max of a sine curve happens at the quarter period and the min happens at the 3/4 period

the max of 325volts occurs at 1/200 sec and the min of -325 occurs at 3/200 sec