In an ac circuit the instantaneous values of current and applied voltage are respectively `i=2(Amp) sin (250pis^(-1)t and epsi =(10V) sin [(250pis^(-1))t+(pi)/(3)`, Find the instantaneous power drawn from the source at `t=(2)/(3)ms` and its average value.

Answer

To solve the problem, we need to find the instantaneous power drawn from the source at \( t = \frac{2}{3} \) ms and its average value. The instantaneous power \( P(t) \) in an AC circuit is given by the product of the instantaneous voltage \( V(t) \) and the instantaneous current \( I(t) \).

### Step 1: Write down the expressions for current and voltage.
The instantaneous current and voltage are given as:
- \( i(t) = 2 \sin(250 \pi t) \) (in Amperes)
- \( e(t) = 10 \sin\left(250 \pi t + \frac{\pi}{3}\right) \) (in Volts)

### Step 2: Substitute \( t = \frac{2}{3} \) ms into the expressions.
First, convert \( t = \frac{2}{3} \) ms to seconds:
\[ t = \frac{2}{3} \times 10^{-3} \text{ s} = \frac{2}{3000} \text{ s} = \frac{1}{1500} \text{ s} \]

Now substitute \( t \) into the expressions for current and voltage.

#### For voltage:
\[ e\left(\frac{2}{3} \text{ ms}\right) = 10 \sin\left(250 \pi \left(\frac{2}{3} \times 10^{-3}\right) + \frac{\pi}{3}\right) \]
Calculating \( 250 \pi \times \frac{2}{3} \times 10^{-3} \):
\[ = \frac{500 \pi}{3} \times 10^{-3} \text{ radians} \]
Thus:
\[ e\left(\frac{2}{3} \text{ ms}\right) = 10 \sin\left(\frac{500 \pi}{3} + \frac{\pi}{3}\right) = 10 \sin\left(\frac{501 \pi}{3}\right) \]

#### For current:
\[ i\left(\frac{2}{3} \text{ ms}\right) = 2 \sin\left(250 \pi \left(\frac{2}{3} \times 10^{-3}\right)\right) = 2 \sin\left(\frac{500 \pi}{3}\right) \]

### Step 3: Calculate the values of voltage and current.
Now we need to evaluate \( \sin\left(\frac{501 \pi}{3}\right) \) and \( \sin\left(\frac{500 \pi}{3}\right) \).

Using the periodic properties of sine:
\[ \frac{501 \pi}{3} = 167 \pi + \frac{2\pi}{3} \Rightarrow \sin\left(\frac{501 \pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
\[ \frac{500 \pi}{3} = 166 \pi + \frac{2\pi}{3} \Rightarrow \sin\left(\frac{500 \pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Substituting these values back:
\[ e\left(\frac{2}{3} \text{ ms}\right) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ V} \]
\[ i\left(\frac{2}{3} \text{ ms}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \text{ A} \]

### Step 4: Calculate the instantaneous power.
Now, we can calculate the instantaneous power:
\[ P(t) = e(t) \cdot i(t) = (5\sqrt{3}) \cdot (\sqrt{3}) = 15 \text{ W} \]

### Step 5: Calculate the average power.
The average power \( P_{avg} \) in an AC circuit is given by:
\[ P_{avg} = \frac{V_{rms} \cdot I_{rms} \cdot \cos(\phi)}{2} \]
Where \( V_{rms} = \frac{V_{max}}{\sqrt{2}} \) and \( I_{rms} = \frac{I_{max}}{\sqrt{2}} \).

Here, \( V_{max} = 10 \text{ V} \) and \( I_{max} = 2 \text{ A} \).
The phase difference \( \phi \) can be determined from the voltage equation:
\[ \phi = \frac{\pi}{3} \]
Thus:
\[ P_{avg} = \frac{10/\sqrt{2} \cdot 2/\sqrt{2} \cdot \cos\left(\frac{\pi}{3}\right)}{2} = \frac{10 \cdot 2 \cdot \frac{1}{2}}{2} = 5 \text{ W} \]

### Final Answers:
- Instantaneous Power at \( t = \frac{2}{3} \) ms: **15 W**
- Average Power: **5 W**