To find the average current over the given time interval for the function $I = 2\sin(60\pi t) + \cos(120\pi t)$, we proceed as follows:

Step-by-Step Solution

1. Define the Average Formula: The average value of a function $I(t)$ over an interval $[a, b]$ is given by: $\text{Average} = \frac{1}{b - a} \int_a^b I(t) \, dt$ In this case: $a = 0 \quad \text{and} \quad b = \frac{1}{600}$

2. Set Up the Integral: Substitute $I(t) = 2\sin(60\pi t) + \cos(120\pi t)$ into the integral: $\text{Average} = \frac{1}{\frac{1}{600} - 0} \int_0^{\frac{1}{600}} \left( 2\sin(60\pi t) + \cos(120\pi t) \right) \, dt$ Simplify the denominator: $\text{Average} = 600 \int_0^{\frac{1}{600}} \left( 2\sin(60\pi t) + \cos(120\pi t) \right) \, dt$

3. Evaluate the Integral: Now, calculate each term in the integral separately: $\int_0^{\frac{1}{600}} 2\sin(60\pi t) \, dt \quad \text{and} \quad \int_0^{\frac{1}{600}} \cos(120\pi t) \, dt$ Each term oscillates over the period, and since the interval $\frac{1}{600}$ matches an integer fraction of the oscillation period for both sine and cosine components, these integrals over the complete cycle average to zero.

4. Conclude: Since both components contribute zero over this interval, the average current is: $\text{Average} = 0$

Therefore, the average current for the given time interval $0 \leq t \leq \frac{1}{600}$ is: $\boxed{0.000}$

Would you like a detailed breakdown of each integral, or do you have any further questions?

Here are some related questions for further exploration:

1. How would the average current change if the interval were different?
2. What is the significance of the period of oscillation in calculating averages?
3. How does the presence of both sine and cosine terms affect the behavior of $I(t)$?
4. Can this method be applied to find the average of any oscillating function?
5. How does this result apply to real-world AC circuits?

Tip: For oscillating functions over an integer fraction of their period, averages often simplify due to symmetry.