The current l 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 dring `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 from \( t = 0 \) to \( t = 2 \) seconds, we start with the given current function:

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

where \( I \) is in amperes and \( t \) is in seconds.

### 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 implies that the differential charge \( dq \) can be expressed as:

\[ dq = I \, dt \]

### Step 2: Substitute the current function
Substituting the expression for \( I(t) \) into the equation for \( dq \):

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

### Step 3: Integrate to find total charge
To find the total charge \( Q \) that passes through the cross-section from \( t = 0 \) to \( t = 2 \) seconds, we need to integrate \( dq \):

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

### Step 4: Calculate the integral
Now, we compute the integral:

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

Calculating each part:

1. For the first integral:

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

2. For the second integral:

\[ \int_{0}^{2} 2t \, dt = t^2 \bigg|_{0}^{2} = (2^2) - (0^2) = 4 \]

### Step 5: Combine the results
Now, we combine the results of the two integrals:

\[ Q = 6 + 4 = 10 \]

### Step 6: State the final answer
Thus, the total charge passed through the cross-section from \( t = 0 \) to \( t = 2 \) seconds is:

\[ Q = 10 \, \text{Coulombs} \]