The charge flowing in a conductor varies with time as `Q= at -(1)/(2)bt^(2)+(1)/(6)ct^(3)` where a, b and c are positive constants. If at time t, the current in the conductor is i, which of the following graphs is correct ?

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the relationship between charge \( Q \) and current \( I \) as given in the equation:

\[ Q = at - \frac{1}{2}bt^2 + \frac{1}{6}ct^3 \]

where \( a \), \( b \), and \( c \) are positive constants. The current \( I \) is defined as the rate of change of charge with respect to time:

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

### Step-by-Step Solution:

1. **Differentiate the Charge Function**:
To find the current \( I \), we need to differentiate \( Q \) with respect to \( t \):
\[ I = \frac{dQ}{dt} = \frac{d}{dt}\left(at - \frac{1}{2}bt^2 + \frac{1}{6}ct^3\right) \]

2. **Apply the Derivative**:
Using basic differentiation rules:
\[ I = a - bt + \frac{1}{2}ct^2 \]

3. **Rearranging the Current Equation**:
We can rearrange the equation for clarity:
\[ I = \frac{1}{2}ct^2 - bt + a \]

4. **Identify the Nature of the Graph**:
The equation \( I = \frac{1}{2}ct^2 - bt + a \) is a quadratic equation in \( t \). Since the coefficient of \( t^2 \) (which is \( \frac{1}{2}c \)) is positive, the graph of \( I \) versus \( t \) will be a parabola that opens upwards.

5. **Find the Minimum Current**:
To find the time at which the current reaches its minimum value, we need to find the vertex of the parabola. The vertex \( t \) can be found using the formula:
\[ t = -\frac{b}{2a} \]
Here, \( a = \frac{1}{2}c \) and \( b = -b \) (the linear coefficient). Thus:
\[ t = \frac{b}{c} \]

6. **Conclusion**:
The minimum current occurs at \( t = \frac{b}{c} \). Therefore, we can check the options provided in the question to identify which graph corresponds to this behavior.