The current in conductor varies with time `t` as `I = 2 t + 3 t^(2)` where `I` is in ampere and `t` in seconds. Electric charge flowing through a section of the conductor during `t = 2 sec` to `t = 3 sec` is

Options

A. 10C
B. 24C
C. 33C
D. 44C

Answer

To find the electric charge flowing through a section of the conductor during the time interval from \( t = 2 \) seconds to \( t = 3 \) seconds, we can follow these steps:

### Step 1: Understand the relationship between current and charge
The current \( I \) in a conductor is defined as the rate of flow of electric charge \( Q \) with respect to time \( t \):
\[ I = \frac{dQ}{dt} \]
This implies that the charge \( dQ \) flowing through the conductor over a small time interval \( dt \) can be expressed as:
\[ dQ = I \, dt \]

### Step 2: Substitute the expression for current
Given the current as a function of time:
\[ I(t) = 2t + 3t^2 \]
we can substitute this into our expression for \( dQ \):
\[ dQ = (2t + 3t^2) \, dt \]

### Step 3: Integrate to find total charge
To find the total charge \( Q \) that flows from \( t = 2 \) seconds to \( t = 3 \) seconds, we need to integrate \( dQ \):
\[ Q = \int_{t=2}^{t=3} (2t + 3t^2) \, dt \]

### Step 4: Perform the integration
We can break this integral into two parts:
\[ Q = \int_{2}^{3} 2t \, dt + \int_{2}^{3} 3t^2 \, dt \]

Calculating the first integral:
\[ \int 2t \, dt = t^2 \quad \text{(evaluated from 2 to 3)} \]
\[ = [3^2 - 2^2] = [9 - 4] = 5 \]

Calculating the second integral:
\[ \int 3t^2 \, dt = t^3 \quad \text{(evaluated from 2 to 3)} \]
\[ = [3^3 - 2^3] = [27 - 8] = 19 \]

### Step 5: Combine the results
Now, we combine the results of the two integrals:
\[ Q = 5 + 19 = 24 \, \text{coulombs} \]

### Final Answer
The electric charge flowing through the section of the conductor during the interval from \( t = 2 \) seconds to \( t = 3 \) seconds is \( \boxed{24} \) coulombs.

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