The charge flowing throug a conductor beginning with time to=0 is given by the formula `q=2t^(2)+3t+1` (coulombs). Find the current `i=(dq)/(dt)` at the end of the 5th seconds.

Answer

To find the current \( i \) at the end of the 5th second, we start with the given formula for the charge \( q \):

\[ q = 2t^2 + 3t + 1 \]

1. **Differentiate the charge with respect to time**: The current \( i \) is defined as the rate of change of charge with respect to time, which can be expressed mathematically as:

\[ i = \frac{dq}{dt} \]

We will differentiate the charge \( q \) with respect to \( t \).

2. **Apply the differentiation**:

- The derivative of \( 2t^2 \) is \( 4t \) (using the power rule).
- The derivative of \( 3t \) is \( 3 \).
- The derivative of the constant \( 1 \) is \( 0 \).

Thus, we have:

\[ \frac{dq}{dt} = 4t + 3 \]

3. **Evaluate the current at \( t = 5 \) seconds**:

Now, we substitute \( t = 5 \) into the expression for current:

\[ i = 4(5) + 3 \]

Calculating this gives:

\[ i = 20 + 3 = 23 \text{ Amperes} \]

4. **Final result**:

The current at the end of the 5th second is:

\[ \boxed{23 \text{ Amperes}} \]

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