The charge flowing in a conductor varies with times as `Q = at - bt^2.` Then, the current

Options

A. reaches a maximum and then decreases
B. falls to zero after `t=(a)/(2b)`
C. changes at a rate of `(-2b)`
D. Both (b) and (c)

Answer

To solve the problem, we need to find the current flowing through a conductor when the charge \( Q \) varies with time as given by the equation:

\[ Q = at - bt^2 \]

where \( a \) and \( b \) are constants.

### Step 1: Find the expression for current \( I \)

The current \( I \) is defined as the rate of change of charge with respect to time. Mathematically, this is expressed as:

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

### Step 2: Differentiate the charge equation

Now, we differentiate the expression for \( Q \):

\[ I = \frac{d}{dt}(at - bt^2) \]

Using the rules of differentiation:

\[ I = a - 2bt \]

### Step 3: Analyze the current expression

The expression \( I = a - 2bt \) indicates that the current is a linear function of time \( t \).

### Step 4: Find the time when current reaches zero

To find when the current reaches zero, we set the expression for current equal to zero:

\[ 0 = a - 2bt \]

Solving for \( t \):

\[ 2bt = a \implies t = \frac{a}{2b} \]

### Step 5: Determine the rate of change of current

Next, we find the rate of change of current by differentiating \( I \):

\[ \frac{dI}{dt} = \frac{d}{dt}(a - 2bt) = -2b \]

### Summary of Results

1. The expression for current is \( I = a - 2bt \).
2. The current reaches zero at \( t = \frac{a}{2b} \).
3. The rate of change of current is \( -2b \).

### Final Answer

Based on the analysis, we conclude:
- The current reaches a maximum and then decreases to zero at \( t = \frac{a}{2b} \).
- The rate of change of current is constant and equal to \( -2b \).