Current flowing through a conducting wire is given by
`I = (1+ 2t)`
Where t is in seconds and current `I` is in amperes. The charge (in coulombs) flown through the resistor in the interval from `t =0` to `t =1` second is -

Options

A. 3
B. 2
C. 1
D. 0

Explanation

To find the charge flown through the resistor from \( t = 0 \) to \( t = 1 \) second, we start with the given current equation:

\[ I = 1 + 2t \]

### Step 1: Relate Current to Charge
We know that current \( I \) is defined as the rate of flow of charge \( Q \) with respect to time \( t \):

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

### Step 2: Substitute the Current Equation
Substituting the expression for current into the equation:

\[ \frac{dQ}{dt} = 1 + 2t \]

### Step 3: Rearrange the Equation
We can rearrange this equation to isolate \( dQ \):

\[ dQ = (1 + 2t) dt \]

### Step 4: Integrate Both Sides
Now, we will integrate both sides. The left side will be integrated with respect to \( Q \) and the right side with respect to \( t \). The limits for \( t \) are from 0 to 1 second, and the initial charge \( Q \) at \( t = 0 \) is 0:

\[ \int_{0}^{Q} dQ = \int_{0}^{1} (1 + 2t) dt \]

### Step 5: Calculate the Integral
Calculating the left side gives:

\[ Q - 0 = \int_{0}^{1} (1 + 2t) dt \]

Now, we compute the right side:

\[ \int (1 + 2t) dt = t + t^2 \]

Evaluating this from 0 to 1:

\[ \left[ t + t^2 \right]_{0}^{1} = (1 + 1^2) - (0 + 0^2) = 1 + 1 - 0 = 2 \]

### Step 6: Final Result
Thus, the charge \( Q \) that has flowed through the resistor from \( t = 0 \) to \( t = 1 \) second is:

\[ Q = 2 \text{ coulombs} \]

### Summary
The charge flown through the resistor in the interval from \( t = 0 \) to \( t = 1 \) second is \( 2 \) coulombs.

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