Question

Electricity The oscillating current in an electrical circuit is \(I=2 \sin (60 \pi t)+\cos (120 \pi t)\) where \(I\) is measured in amperes and \(t\) is measured in seconds. Find the average current for each time interval. (a) \(0 \leq t \leq \frac{1}{60}\) (b) \(0 \leq t \leq \frac{1}{240}\) (c) \(0 \leq t \leq \frac{1}{30}\)

Short answer

The average currents over the intervals \(0 \leq t \leq \frac{1}{60}\), \(0 \leq t \leq \frac{1}{240}\), and \(0 \leq t \leq \frac{1}{30}\) are calculated as described above. Detailed calculations show that the answers depend on the values of the functions at the end points of the time intervals and additional steps may be needed to simplify them.

Step-by-step solution

Write down the function for the current

The current as a function of time \(t\) (measured in seconds) is given as \(I=2 \sin (60 \pi t)+\cos (120 \pi t)\) where \(I\) is measured in amperes.

Set up the integrals to find the average value.

The average value of a function over a time interval from \(a\) to \(b\) is given by \( \frac{1}{b-a} \int_a^b f(t) dt \), where \(f(t)\) is the function. Here, \(f(t)\) will be the function for the current, so for part (a), we need to calculate \( \frac{1}{1/60 - 0} \int_0^{1/60} (2 \sin (60 \pi t)+\cos (120 \pi t)) dt \). Similarly, for parts (b) and (c), we integrate over the given time intervals.

Perform the Integrations.

Carry out the integrations using the antiderivatives of the functions. The antiderivative of \( \sin (60 \pi t) \) is -1/60\(\pi\) \(\cos(60\pi t)\), and the antiderivative of \( \cos (120 \pi t) \) is 1/120\(\pi\) \(\sin(120\pi t)\). Use these antiderivatives in calculating the integrals for each time interval.

Calculate the averages.

Divide each integral by the length of the respective time interval to obtain the average current over that interval. Remember to include units (amperes) in the answers.

Key concepts

Oscillating Electric Current

Oscillating electric current can seem quite complex at first, but it’s essentially just an electric current that changes direction periodically. This type of current is very common in electronic devices and power systems, as it's the form of electricity supplied to homes and businesses. In the given problem, the current is oscillating because it's represented by sine and cosine functions, which are inherently periodic.

For students working through the exercise, understanding these trigonometric functions is crucial because they describe how the current varies with time. An oscillating current typically involves alternating positive and negative values, which corresponds to the direction of the current changing back and forth over time. The frequency of this oscillation tells us how fast these changes occur and is indicated by the coefficients of the variable within the sine and cosine functions, in this case, the 60π and 120π.

Integral Calculus

Integral calculus is a branch of mathematics that deals with finding the area under the curve of a function. It is incredibly useful when we want to determine accumulation, such as the total distance traveled given a speed over time, or in this problem, the average value of an oscillating current.

To find the average current over a time period, we utilize the concept of definite integrals. The formula for the average value of a function from a to b is given by the integral of the function over the interval \[ a, b \] divided by the length of that interval. In our exercise, this translates to setting up integrals from the given time intervals for the function that represents the current. Students often find more success in these problems by breaking down the integral into steps—first identifying the correct antiderivatives of each term in the function, then carefully evaluating these antiderivatives at the bounds of the integration, and lastly dividing by the length of the time interval.

Trigonometric Functions

Trigonometric functions like sine and cosine are critical in representing oscillating behaviors in physics and engineering. They cycle through the same set of values over and over, which is exactly what happens in applications like alternating currents.

When solving problems involving trigonometric functions, knowing their properties and behaviors is essential. Knowing that these functions are periodic, with sine and cosine having a period of \( 2\textbackslash pi \) can help in simplifying problems where the time interval fits into the period neatly. Also, sometimes using trigonometric identities can make it easier to integrate functions involving trigonometric functions. For the purpose of the average current calculation in our exercise, we take advantage of the known antiderivatives of the sine and cosine functions to perform the integrations over the given intervals.