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.

Explanation

To solve the problem, we need to find the instantaneous power drawn from the source at \( t = \frac{2}{3} \, \text{ms} \) and its average value.

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

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

Now substitute \( t \) into the current and voltage equations:
- For current:
\[ i\left(\frac{2}{3} \times 10^{-3}\right) = 2 \sin\left(250 \pi \times \frac{2}{3} \times 10^{-3}\right) = 2 \sin\left(\frac{500 \pi}{3} \times 10^{-3}\right) = 2 \sin\left(\frac{500 \pi}{3} \times \frac{1}{1000}\right) = 2 \sin\left(\frac{500 \pi}{3000}\right) \]

- For voltage:
\[ e\left(\frac{2}{3} \times 10^{-3}\right) = 10 \sin\left(250 \pi \times \frac{2}{3} \times 10^{-3} + \frac{\pi}{3}\right) = 10 \sin\left(\frac{500 \pi}{3} \times 10^{-3} + \frac{\pi}{3}\right) \]

### Step 3: Calculate the values of current and voltage
Now we can simplify:
- Current:
\[ i\left(\frac{2}{3} \times 10^{-3}\right) = 2 \sin\left(\frac{500 \pi}{3000}\right) = 2 \sin\left(\frac{5 \pi}{30}\right) = 2 \sin\left(\frac{\pi}{6}\right) = 2 \times \frac{1}{2} = 1 \, \text{Amp} \]

- Voltage:
\[ e\left(\frac{2}{3} \times 10^{-3}\right) = 10 \sin\left(\frac{500 \pi}{3000} + \frac{\pi}{3}\right) = 10 \sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = 10 \sin\left(\frac{\pi}{6} + \frac{2\pi}{6}\right) = 10 \sin\left(\frac{3\pi}{6}\right) = 10 \sin\left(\frac{\pi}{2}\right) = 10 \times 1 = 10 \, \text{Volts} \]

### Step 4: Calculate the instantaneous power
The instantaneous power \( P \) is given by:
\[ P = V \cdot I = e(t) \cdot i(t) = 10 \cdot 1 = 10 \, \text{Watts} \]

### Step 5: Calculate the average power
For AC circuits, the average power \( P_{\text{avg}} \) can be calculated using:
\[ P_{\text{avg}} = \frac{V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi}{1} \]
Where \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \) and \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \).

Given:
- \( V_0 = 10 \, \text{V} \)
- \( I_0 = 2 \, \text{A} \)
- Phase difference \( \phi = \frac{\pi}{3} \)

Calculating:
\[ V_{\text{rms}} = \frac{10}{\sqrt{2}} \approx 7.07 \, \text{V} \]
\[ I_{\text{rms}} = \frac{2}{\sqrt{2}} \approx 1.41 \, \text{A} \]
\[ P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos\left(\frac{\pi}{3}\right) = 7.07 \cdot 1.41 \cdot \frac{1}{2} \approx 5 \, \text{Watts} \]

### Final Answers:
- Instantaneous Power at \( t = \frac{2}{3} \, \text{ms} \): \( 10 \, \text{Watts} \)
- Average Power: \( 5 \, \text{Watts} \)