If charge flown through a wire is given by `q=3sin(3t)`, then
Q. Find out the current flown through the wire at `t=(pi)/(9)` second.

Options

A. 4.5 amp
B. `4.5sqrt3amp`
C. `(sqrt3)/(2)amp`
D. `9amp`

Explanation

To find the current flowing through the wire at time \( t = \frac{\pi}{9} \) seconds, we start with the given expression for charge:

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

### Step 1: Find the expression for current
The current \( I \) is defined as the rate of change of charge with respect to time, which can be expressed mathematically as:

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

### Step 2: Differentiate the charge function
We need to differentiate \( q(t) \) with respect to \( t \):

\[ I = \frac{d}{dt}(3 \sin(3t)) \]

Using the chain rule, we differentiate:

\[ I = 3 \cdot \cos(3t) \cdot \frac{d}{dt}(3t) = 3 \cdot \cos(3t) \cdot 3 = 9 \cos(3t) \]

So, the expression for current is:

\[ I(t) = 9 \cos(3t) \]

### Step 3: Substitute \( t = \frac{\pi}{9} \)
Now we need to find the current at \( t = \frac{\pi}{9} \):

\[ I\left(\frac{\pi}{9}\right) = 9 \cos\left(3 \cdot \frac{\pi}{9}\right) \]

This simplifies to:

\[ I\left(\frac{\pi}{9}\right) = 9 \cos\left(\frac{\pi}{3}\right) \]

### Step 4: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
We know that:

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 5: Final calculation
Substituting this value back into the expression for current:

\[ I\left(\frac{\pi}{9}\right) = 9 \cdot \frac{1}{2} = 4.5 \text{ A} \]

Thus, the current flowing through the wire at \( t = \frac{\pi}{9} \) seconds is:

\[ \boxed{4.5 \text{ A}} \]

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