The current I through a given cross -section varies with time t as `I = 3+ 2t ` , where l is in ampere and t is in second. The charge passed through this cross-section during `t=0` to `t=2` s is

Options

A. 7C
B. 10C
C. 14C
D. 20C

Answer

To find the charge passed through a cross-section when the current varies with time, we can use the relationship between current (I) and charge (Q). The current is defined as the rate of flow of charge, which can be expressed mathematically as:

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

Given that the current varies with time as:

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

where \( I \) is in amperes and \( t \) is in seconds, we can find the total charge \( Q \) that passes through the cross-section from \( t = 0 \) to \( t = 2 \) seconds by integrating the current over this time interval.

### Step-by-Step Solution:

1. **Set up the integral for charge**:
Since \( I = \frac{dQ}{dt} \), we can express the charge \( Q \) as:
\[ Q = \int_{t_1}^{t_2} I(t) \, dt \]
Here, \( t_1 = 0 \) and \( t_2 = 2 \).

2. **Substitute the expression for current**:
Substitute \( I(t) = 3 + 2t \) into the integral:
\[ Q = \int_{0}^{2} (3 + 2t) \, dt \]

3. **Evaluate the integral**:
We can break this integral into two parts:
\[ Q = \int_{0}^{2} 3 \, dt + \int_{0}^{2} 2t \, dt \]

- The first integral:
\[ \int_{0}^{2} 3 \, dt = 3t \bigg|_{0}^{2} = 3(2) - 3(0) = 6 \]

- The second integral:
\[ \int_{0}^{2} 2t \, dt = 2 \cdot \frac{t^2}{2} \bigg|_{0}^{2} = t^2 \bigg|_{0}^{2} = (2^2) - (0^2) = 4 \]

4. **Combine the results**:
Now, add the results of the two integrals:
\[ Q = 6 + 4 = 10 \, \text{coulombs} \]

### Final Answer:
The total charge passed through the cross-section during the time interval from \( t = 0 \) to \( t = 2 \) seconds is \( 10 \, \text{coulombs} \).