The charge flowing in a conductor varies with time so `q = 2t - 6t^(2)+10t^(3)`
where `q`is in coloumn and `r` in second Find (i) the initial current (ii) the time after which the value of current reaches is maximum value (ii) the maximum or minimum value of current

Explanation

To solve the problem, we will follow these steps:

### Given:
The charge flowing in a conductor is given by:
\[ q(t) = 2t - 6t^2 + 10t^3 \]
where \( q \) is in coulombs and \( t \) is in seconds.

### Step 1: Find the expression for current \( I(t) \)
Current \( I(t) \) is defined as the rate of change of charge with respect to time:
\[ I(t) = \frac{dq}{dt} \]

Calculating the derivative:
\[ I(t) = \frac{d}{dt}(2t - 6t^2 + 10t^3) = 2 - 12t + 30t^2 \]

### Step 2: Find the initial current \( I(0) \)
To find the initial current, we substitute \( t = 0 \) into the current equation:
\[ I(0) = 2 - 12(0) + 30(0)^2 = 2 \text{ Amperes} \]

### Step 3: Find the time when current reaches its maximum value
To find the time at which the current is maximum, we need to set the derivative of the current \( I(t) \) to zero:
\[ \frac{dI}{dt} = 0 \]
Calculating the derivative of \( I(t) \):
\[ \frac{dI}{dt} = -12 + 60t \]
Setting this equal to zero:
\[ -12 + 60t = 0 \implies 60t = 12 \implies t = \frac{12}{60} = \frac{1}{5} \text{ seconds} \]

### Step 4: Find the maximum value of current \( I(t) \)
Now, we substitute \( t = \frac{1}{5} \) back into the current equation to find the maximum current:
\[ I\left(\frac{1}{5}\right) = 2 - 12\left(\frac{1}{5}\right) + 30\left(\frac{1}{5}\right)^2 \]
Calculating each term:
\[ I\left(\frac{1}{5}\right) = 2 - \frac{12}{5} + 30 \cdot \frac{1}{25} \]
\[ = 2 - \frac{12}{5} + \frac{30}{25} = 2 - \frac{12}{5} + \frac{6}{5} = 2 - \frac{6}{5} = \frac{10}{5} - \frac{6}{5} = \frac{4}{5} = 0.8 \text{ Amperes} \]

### Summary of Results:
1. Initial current \( I(0) = 2 \) Amperes.
2. Time when current reaches maximum \( t = \frac{1}{5} \) seconds.
3. Maximum current value \( I\left(\frac{1}{5}\right) = 0.8 \) Amperes.