The current in a conductor varies with time 't' as I = 3t + 4`t^2`. Where I in amp and t in sec. The electric charge flows through the section of the conductor between t = 1s and t = 3s

Options

A. `(14)/(3) C `
B. `(3)/(14) C`
C. `(140)/(3) C `
D. `(3)/( 140)C `

Answer

To find the electric charge that flows through a section of the conductor between \( t = 1 \) s and \( t = 3 \) s, we will use the relationship between current and charge. The current \( I \) is given by the equation:

\[ I = 3t + 4t^2 \]

We know that current is the rate of flow of charge, which can be expressed mathematically as:

\[ Q = \int_{t_1}^{t_2} I \, dt \]

Where:
- \( Q \) is the total charge,
- \( I \) is the current,
- \( t_1 \) and \( t_2 \) are the time limits (in this case, \( t_1 = 1 \) s and \( t_2 = 3 \) s).

### Step 1: Set up the integral for charge
We will set up the integral to find the charge \( Q \):

\[ Q = \int_{1}^{3} (3t + 4t^2) \, dt \]

### Step 2: Break down the integral
We can break this integral into two parts:

\[ Q = \int_{1}^{3} 3t \, dt + \int_{1}^{3} 4t^2 \, dt \]

### Step 3: Calculate the first integral
Now we calculate the first integral:

\[ \int 3t \, dt = \frac{3t^2}{2} \]

Evaluating from \( t = 1 \) to \( t = 3 \):

\[ \left[ \frac{3(3)^2}{2} - \frac{3(1)^2}{2} \right] = \left[ \frac{3 \cdot 9}{2} - \frac{3 \cdot 1}{2} \right] = \left[ \frac{27}{2} - \frac{3}{2} \right] = \frac{24}{2} = 12 \]

### Step 4: Calculate the second integral
Now we calculate the second integral:

\[ \int 4t^2 \, dt = \frac{4t^3}{3} \]

Evaluating from \( t = 1 \) to \( t = 3 \):

\[ \left[ \frac{4(3)^3}{3} - \frac{4(1)^3}{3} \right] = \left[ \frac{4 \cdot 27}{3} - \frac{4 \cdot 1}{3} \right] = \left[ \frac{108}{3} - \frac{4}{3} \right] = \frac{104}{3} \]

### Step 5: Combine the results
Now we combine the results from both integrals to find the total charge \( Q \):

\[ Q = 12 + \frac{104}{3} \]

To add these, we convert \( 12 \) into a fraction with a denominator of \( 3 \):

\[ 12 = \frac{36}{3} \]

So,

\[ Q = \frac{36}{3} + \frac{104}{3} = \frac{140}{3} \]

### Final Answer
Thus, the total charge that flows through the section of the conductor between \( t = 1 \) s and \( t = 3 \) s is:

\[ Q = \frac{140}{3} \text{ coulombs} \]