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 in the interval 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 expression for current
Substituting the expression for \( I \) into the equation, we have:

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

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

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

### Step 4: Integrate both sides
Now, we will integrate both sides. We will integrate \( dQ \) from \( 0 \) to \( Q \) and \( dt \) from \( 0 \) to \( 1 \):

\[ \int_0^Q dQ = \int_0^1 (1 + 2t) dt \]

### Step 5: Solve the right-hand side integral
Calculating the right-hand side:

\[ \int_0^1 (1 + 2t) dt = \int_0^1 1 \, dt + \int_0^1 2t \, dt \]

Calculating each part:

1. \(\int_0^1 1 \, dt = [t]_0^1 = 1 - 0 = 1\)
2. \(\int_0^1 2t \, dt = [t^2]_0^1 = 1^2 - 0^2 = 1\)

Thus, the right-hand side becomes:

\[ 1 + 1 = 2 \]

### Step 6: Set the integrals equal
Now we equate the results of the integrals:

\[ Q = 2 \]

### Conclusion
The charge flown through the resistor in the interval from \( t = 0 \) to \( t = 1 \) second is:

\[ \boxed{2 \text{ Coulombs}} \]

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