The current in a wire varies with time according to the equation l=4+2t, where l is in ampere and t is in sec. the quantity of charge which has passed through a cross-section of the wire during the time t=2 sec to t=6 sec will be

Options

A. 60 coulomb
B. 24 coulomb
C. 48 coulomb
D. 30 coulomb

Explanation

To find the quantity of charge that has passed through a cross-section of the wire from \( t = 2 \) seconds to \( t = 6 \) seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between current and charge**:
The current \( I \) in the wire is given by the equation:
\[ I = 4 + 2t \]
The current is defined as the rate of flow of charge, mathematically expressed as:
\[ I = \frac{dq}{dt} \]
where \( dq \) is the differential charge and \( dt \) is the differential time.

2. **Rearrange the equation**:
From the current equation, we can express \( dq \) as:
\[ dq = (4 + 2t) dt \]

3. **Integrate to find the total charge**:
To find the total charge \( Q \) that flows from \( t = 2 \) seconds to \( t = 6 \) seconds, we need to integrate \( dq \) from \( t = 2 \) to \( t = 6 \):
\[ Q = \int_{2}^{6} (4 + 2t) dt \]

4. **Calculate the integral**:
We can break the integral into two parts:
\[ Q = \int_{2}^{6} 4 dt + \int_{2}^{6} 2t dt \]
- For the first integral:
\[ \int_{2}^{6} 4 dt = 4[t]_{2}^{6} = 4(6 - 2) = 4 \times 4 = 16 \]
- For the second integral:
\[ \int_{2}^{6} 2t dt = 2\left[\frac{t^2}{2}\right]_{2}^{6} = [t^2]_{2}^{6} = (6^2 - 2^2) = 36 - 4 = 32 \]

5. **Combine the results**:
Now we can combine the results of both integrals:
\[ Q = 16 + 32 = 48 \text{ Coulombs} \]

### Final Answer:
The quantity of charge that has passed through the cross-section of the wire from \( t = 2 \) seconds to \( t = 6 \) seconds is **48 Coulombs**.