A current through a wire depends on time as `i = alpha_0 t + beta t^2 ` where `alpha_0 = 20 A//s ` and `beta = 8As^(-2)` . Find the charge crossed through a section of the wire in 15 s.

Options

A. 2250 C
B. 11250 C
C. 2100 C
D. 260 C

Answer

To find the total charge that has crossed through a section of the wire in 15 seconds, we start with the given current equation:

\[ i(t) = \alpha_0 t + \beta t^2 \]

where:
- \( \alpha_0 = 20 \, \text{A/s} \)
- \( \beta = 8 \, \text{A/s}^2 \)

The charge \( Q \) that flows through the wire over a time interval can be calculated by integrating the current over that time interval:

\[ Q = \int_{0}^{t} i(t) \, dt \]

### Step 1: Substitute the expression for current into the integral

Substituting \( i(t) \) into the integral gives:

\[ Q = \int_{0}^{15} (\alpha_0 t + \beta t^2) \, dt \]

### Step 2: Substitute the values of \( \alpha_0 \) and \( \beta \)

Now substituting the values of \( \alpha_0 \) and \( \beta \):

\[ Q = \int_{0}^{15} (20t + 8t^2) \, dt \]

### Step 3: Calculate the integral

We can now calculate the integral:

1. The integral of \( 20t \) is \( 10t^2 \).
2. The integral of \( 8t^2 \) is \( \frac{8}{3}t^3 \).

So we have:

\[ Q = \left[ 10t^2 + \frac{8}{3}t^3 \right]_{0}^{15} \]

### Step 4: Evaluate the integral at the limits

Now we evaluate this expression at the upper limit \( t = 15 \) and the lower limit \( t = 0 \):

1. At \( t = 15 \):

\[ Q = 10(15^2) + \frac{8}{3}(15^3) \]

Calculating each term:

- \( 10(15^2) = 10 \times 225 = 2250 \)
- \( \frac{8}{3}(15^3) = \frac{8}{3} \times 3375 = 9000 \)

So:

\[ Q = 2250 + 9000 = 11250 \]

2. At \( t = 0 \):

\[ Q = 10(0^2) + \frac{8}{3}(0^3) = 0 \]

### Step 5: Final result

Thus, the total charge \( Q \) that has crossed through the section of the wire in 15 seconds is:

\[ Q = 11250 \, \text{C} \]