The voltage of the power supply in a AC-DC coupled circuit is given by the function v(t) = 130sin5t + 18, where t is time, in seconds, and V is the voltage, in volts, at time t.

a) find the maxiumum and mimimum voltages and the times at which they occur

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The max/min can actually be calculated quite easily by just noting the property of the sine and cosine curves

No matter what the input, sinx has a max of +1 and a min of -1
so sin(5t) has a max of 1 and a min of -1
130sin(5t) has a max of 130 and a min of -130
130sin(5t) + 18 has a max of 148 and a min of -112

To find the maximum and minimum voltages, we need to first determine the amplitude of the sine function. The amplitude is the coefficient of the sine function, which is 130 in this case.

a) Maximum voltage:
The maximum voltage occurs when the sine function has a value of +1. Therefore, the maximum voltage is given by adding the amplitude to the constant term:
Maximum voltage = 130 + 18 = 148 volts

To find the time at which the maximum voltage occurs, we need to find the value of t for which the sine function reaches its maximum value of +1. The maximum value of the sine function occurs when the argument of the sine function, 5t, is equal to π/2 (since sin(π/2) = 1). Therefore, we can solve for t as follows:

5t = π/2
t = (π/2) / 5
t ≈ 0.314 seconds

So, the maximum voltage of 148V occurs at approximately 0.314 seconds.

b) Minimum voltage:
The minimum voltage occurs when the sine function has a value of -1. Therefore, the minimum voltage is given by subtracting the amplitude from the constant term:
Minimum voltage = -130 + 18 = -112 volts

Similarly, to find the time at which the minimum voltage occurs, we need to find the value of t for which the sine function reaches its minimum value of -1. This occurs when the argument of the sine function, 5t, is equal to 3π/2 (since sin(3π/2) = -1). Solving for t:

5t = 3π/2
t = (3π/2) / 5
t ≈ 0.942 seconds

Therefore, the minimum voltage of -112V occurs at approximately 0.942 seconds.

To find the maximum and minimum voltages of the function v(t) = 130sin(5t) + 18, we can observe that the maximum and minimum values of the sine function occur at specific values of the angle within one period of the function.

For clarity, let's rewrite the function as v(t) = 130sin(5t) + 18.

1. Maximum Voltage:
The maximum value of the sine function is 1. Therefore, to find the maximum voltage, we need to find the value of t that makes sin(5t) equal to 1, and then substitute that value into the function v(t).

sin(5t) = 1
To solve for t, we can take the inverse sine (arcsin) of both sides:
5t = arcsin(1)
5t = π/2

Dividing both sides by 5, we get:
t = (π/2) / 5
t = π/10

Now, substitute t = π/10 into the function v(t):
v(π/10) = 130sin(5(π/10)) + 18
v(π/10) = 130sin(π/2) + 18
v(π/10) = 130 + 18
v(π/10) = 148

Therefore, the maximum voltage is 148 volts, and it occurs at t = π/10 seconds.

2. Minimum Voltage:
The minimum value of the sine function is -1. Therefore, to find the minimum voltage, we need to find the value of t that makes sin(5t) equal to -1, and then substitute that value into the function v(t).

sin(5t) = -1
To solve for t, we can take the inverse sine (arcsin) of both sides:
5t = arcsin(-1)
5t = -π/2

Dividing both sides by 5, we get:
t = (-π/2) / 5
t = -π/10

Now, substitute t = -π/10 into the function v(t):
v(-π/10) = 130sin(5(-π/10)) + 18
v(-π/10) = 130sin(-π/2) + 18
v(-π/10) = -130 + 18
v(-π/10) = -112

Therefore, the minimum voltage is -112 volts, and it occurs at t = -π/10 seconds.