Question

The electric current through a wire varies with time as $I = I_0 + \beta t$ where $I_0 = 20 \, \text{A}$ and $\beta = 3 \, \text{A/s}$. The amount of electric charge that crosses through a section of the wire in $20 \, \text{s}$ is:

• A. 80C
• B. 1000C
• C. 800C
• D. 1600C

Answer 1

Answer: (B)

1000C

Answer 2

Step 1: Express Current as a Function of Time

$I = I_0 + \beta t = 20 + 3t$

Step 2: Calculate Charge $q$ Over Time

The current $I = \frac{dq}{dt}$, so we can write:

$dq = (20 + 3t) dt$

Step 3: Integrate to Find Total Charge $q$ from $t = 0$ to $t = 20$

$q = \int_{0}^{20} (20 + 3t) dt$

Split the integral:

$q = \int_{0}^{20} 20 dt + \int_{0}^{20} 3t dt$

Step 4: Evaluate Each Integral

$q = \left[20t\right]_{0}^{20} + \left[\frac{3t^2}{2}\right]_{0}^{20}$

$= (20 \times 20) + \frac{3 \times 20^2}{2}$

$= 400 + \frac{3 \times 400}{2}$

$= 400 + 600 = 1000 \, C$

So, the correct answer is: 1000 C