If charge flown through a cross section of wire in one direction `0` to `t` is given by `q=3 sin(3t)` then
Find out the area under `i-t` curve from `t=(pi)/(9)` to `t=(pi)/(6)` seconds `:`

Options

A. `3[(2-sqrt(3))/(2)]`
B. `3[(2+sqrt(3))/(2)]`
C. `[(2-sqrt(3))/(2)]`
D. `[(2+sqrt(3))/(2)]`

Explanation

To solve the problem, we need to find the area under the current-time (i-t) curve from \( t = \frac{\pi}{9} \) to \( t = \frac{\pi}{6} \) seconds. The charge \( q \) flowing through a cross-section of the wire is given by the equation:

\[ q = 3 \sin(3t) \]

### Step-by-Step Solution:

1. **Understand the relationship between charge and current**:
The current \( i \) is defined as the rate of change of charge with respect to time, which can be expressed as:
\[ i = \frac{dq}{dt} \]
Therefore, the differential charge \( dq \) can be expressed as:
\[ dq = i \, dt \]

2. **Finding the area under the i-t curve**:
The area under the i-t curve from \( t = \frac{\pi}{9} \) to \( t = \frac{\pi}{6} \) is given by:
\[ A = \int_{t_1}^{t_2} i \, dt = \int_{\frac{\pi}{9}}^{\frac{\pi}{6}} i \, dt \]
Since \( i = \frac{dq}{dt} \), we can rewrite the integral as:
\[ A = \int_{q(t_1)}^{q(t_2)} dq = q(t_2) - q(t_1) \]

3. **Calculate \( q(t) \) at the limits**:
We need to evaluate \( q(t) \) at \( t = \frac{\pi}{6} \) and \( t = \frac{\pi}{9} \):
- For \( t = \frac{\pi}{6} \):
\[ q\left(\frac{\pi}{6}\right) = 3 \sin\left(3 \cdot \frac{\pi}{6}\right) = 3 \sin\left(\frac{\pi}{2}\right) = 3 \cdot 1 = 3 \]
- For \( t = \frac{\pi}{9} \):
\[ q\left(\frac{\pi}{9}\right) = 3 \sin\left(3 \cdot \frac{\pi}{9}\right) = 3 \sin\left(\frac{\pi}{3}\right) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

4. **Calculate the area \( A \)**:
Now substituting the values of \( q(t_2) \) and \( q(t_1) \):
\[ A = q\left(\frac{\pi}{6}\right) - q\left(\frac{\pi}{9}\right) = 3 - \frac{3\sqrt{3}}{2} \]
Simplifying this gives:
\[ A = 3 - \frac{3\sqrt{3}}{2} = \frac{6}{2} - \frac{3\sqrt{3}}{2} = \frac{6 - 3\sqrt{3}}{2} \]

5. **Final result**:
The area under the i-t curve from \( t = \frac{\pi}{9} \) to \( t = \frac{\pi}{6} \) is:
\[ A = \frac{3}{2} (2 - \sqrt{3}) \]