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`

Explanation

To find the charge flowing through the cross-section of the wire from time \( t = 0 \) to \( t = 2 \) seconds, we can follow these steps:

### Step 1: Understand the relationship between current and charge
The current \( I \) is defined as the rate of flow of charge, which can be expressed mathematically as:
\[ I = \frac{dq}{dt} \]
where \( dq \) is the infinitesimal charge and \( dt \) is the infinitesimal time interval.

### Step 2: Write the expression for charge in terms of current
Given the current as a function of time:
\[ I = 3t^2 + 2t + 5 \]
we can express \( dq \) as:
\[ dq = (3t^2 + 2t + 5) dt \]

### Step 3: Integrate to find the 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
Now, we can integrate each term separately:
\[ Q = \int_{0}^{2} 3t^2 dt + \int_{0}^{2} 2t dt + \int_{0}^{2} 5 dt \]

Calculating each integral:
1. \(\int 3t^2 dt = t^3\) evaluated from 0 to 2:
\[ = [2^3 - 0^3] = 8 \]

2. \(\int 2t dt = t^2\) evaluated from 0 to 2:
\[ = [2^2 - 0^2] = 4 \]

3. \(\int 5 dt = 5t\) evaluated from 0 to 2:
\[ = [5 \cdot 2 - 5 \cdot 0] = 10 \]

### Step 5: Sum the results of the integrals
Now, summing these results gives:
\[ Q = 8 + 4 + 10 = 22 \text{ coulombs} \]

### Final Answer
The charge flowing through the cross-section of the wire from \( t = 0 \) to \( t = 2 \) seconds is \( 22 \) coulombs.

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