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

The Correct Option is B

To find the amount of electric charge that crosses through the wire in 20 seconds, we start with the relation given for current:

\(I = I_0 + \beta t\) 

where \( I_0 = 20 \, \text{A} \) and \( \beta = 3 \, \text{A/s} \).

The electric charge \( Q \) that passes through the wire can be calculated using the formula:

\(Q = \int I \, dt\)

Substitute the expression for \( I \) into the integral:

\(Q = \int (I_0 + \beta t) \, dt\)

Evaluate the integral over the time interval from \( t = 0 \) to \( t = 20 \, \text{s} \):

\(Q = \int_0^{20} (20 + 3t) \, dt = [20t + \frac{3}{2}t^2]_0^{20}\)

Calculate the definite integral:

\(= \left( 20 \times 20 + \frac{3}{2} \times 20^2 \right) - \left( 20 \times 0 + \frac{3}{2} \times 0^2 \right)\)

\(= 400 + \frac{3}{2} \times 400\)

\(= 400 + 600\)

\(= 1000 \, \text{C}\)

Hence, the correct amount of electric charge that crosses through the wire in 20 seconds is 1000 C.

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