The current flowing through wire depends on time as, `I = 3t^(2) + 2t + 5`
The charge flowing through the cross - section of the wire in time `t = 0` to `t = 2` second is

Options

A. 22 C
B. 20 C
C. 18 C
D. 5 C

Answer

To find the charge flowing through the cross-section of the wire from time \( t = 0 \) to \( t = 2 \) seconds, we start with the given current function:

\[ I(t) = 3t^2 + 2t + 5 \]

### Step 1: Relate current to charge
The current \( I \) is defined as the rate of flow of charge, which can be expressed mathematically as:

\[ I = \frac{dq}{dt} \]

This means that the differential charge \( dq \) can be expressed as:

\[ dq = I \, dt \]

### Step 2: Substitute the expression for current
Substituting the expression for current into the equation gives:

\[ dq = (3t^2 + 2t + 5) \, dt \]

### Step 3: Integrate to find total charge
To find the total charge \( Q \) that flows from \( t = 0 \) to \( t = 2 \), we need to integrate \( dq \):

\[ Q = \int_{0}^{2} (3t^2 + 2t + 5) \, dt \]

### Step 4: Perform the integration
We can break this integral into three parts:

\[ Q = \int_{0}^{2} 3t^2 \, dt + \int_{0}^{2} 2t \, dt + \int_{0}^{2} 5 \, dt \]

Calculating each integral separately:

1. For \( \int 3t^2 \, dt \):
\[ \int 3t^2 \, dt = 3 \cdot \frac{t^3}{3} = t^3 \quad \text{(evaluated from 0 to 2)} \]
\[ = 2^3 - 0^3 = 8 \]

2. For \( \int 2t \, dt \):
\[ \int 2t \, dt = 2 \cdot \frac{t^2}{2} = t^2 \quad \text{(evaluated from 0 to 2)} \]
\[ = 2^2 - 0^2 = 4 \]

3. For \( \int 5 \, dt \):
\[ \int 5 \, dt = 5t \quad \text{(evaluated from 0 to 2)} \]
\[ = 5 \cdot 2 - 5 \cdot 0 = 10 \]

### Step 5: Combine the results
Now, we combine the results from the three integrals:

\[ Q = 8 + 4 + 10 = 22 \, \text{coulombs} \]

### Conclusion
The total charge flowing through the cross-section of the wire from \( t = 0 \) to \( t = 2 \) seconds is:

\[ \boxed{22 \, \text{coulombs}} \]