Question

The charge flowing in a conductor varies with time as $q = 2t - 6t^2 + 10^3$ where 'q' is in coulomb and 't' in second. find ① Initial current. ② The maximum or minimum value of current. ③ The time after which the current changes, it's direction.

Answer 1

1. Initial Current: $2 \, \text{A}$

2. Maximum Current: $2 \, \text{A}$ (at $t=0$)

3. Time of Direction Change: $t=\frac{1}{6}$ s

Answer 2

Solution

Given:

$$q = 2t - 6t^2 + 10^3 \quad \text{(in coulombs)}$$

Step 1: Find the current $i$

$$i = \frac{dq}{dt} = \frac{d}{dt}\left(2t - 6t^2 + 10^3\right) = 2 - 12t$$

Step 2: Initial Current (at $t=0$)

$$i(0) = 2 - 12(0) = 2 \, \text{A}$$

Step 3: Maximum or Minimum Value of Current

The current $i(t) = 2 - 12t$ is a linear function. For $t \ge 0$, the maximum occurs at $t=0$ and is:

$$\text{Maximum current} = 2 \, \text{A}$$

There is no turning point in this linear function.

Step 4: Time When the Current Changes Direction

The current changes direction when $i = 0$:

$$2 - 12t = 0 \quad \Rightarrow \quad 12t = 2 \quad \Rightarrow \quad t = \frac{2}{12} = \frac{1}{6} \, \text{s}$$

Core Explanation (minimal)

Differentiate $q$ to get $i = 2-12t$. Initial current at $t=0$ is $i(0)=2\,A$. Current becomes zero when $2-12t=0$ i.e. $t=\frac{1}{6}$ s, which is the time after which the current reverses its direction. Since $i(t)$ is linear, the maximum (for $t\ge0$) is $2\,A$ at $t=0$.