In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t
`i = 20 sin( 30t-(pi)/(4))` In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

Options

A. `(50)/(sqrt2),0`
B. 50,0
C. 50,10
D. `(1000)/(sqrt2),10`

Answer

To solve the problem, we need to find the average power consumed by the circuit and the wattless current using the given instantaneous e.m.f. and current equations.

### Step 1: Identify the given equations
The instantaneous e.m.f. and current are given as:
- \( e = 100 \sin(30t) \)
- \( i = 20 \sin(30t - \frac{\pi}{4}) \)

### Step 2: Determine the peak values
From the equations, we can identify the peak values:
- The peak e.m.f. \( E_0 = 100 \, \text{V} \)
- The peak current \( I_0 = 20 \, \text{A} \)

### Step 3: Calculate the RMS values
The RMS (Root Mean Square) values are calculated using the formula:
\[ V_{\text{rms}} = \frac{E_0}{\sqrt{2}} \quad \text{and} \quad I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]
Calculating these:
\[ V_{\text{rms}} = \frac{100}{\sqrt{2}} = 50\sqrt{2} \, \text{V} \]
\[ I_{\text{rms}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{A} \]

### Step 4: Determine the phase difference
The phase difference \( \phi \) can be determined from the current equation:
\[ i = 20 \sin(30t - \frac{\pi}{4}) \]
This indicates that \( \phi = \frac{\pi}{4} \).

### Step 5: Calculate the average power consumed
The average power \( P_{\text{avg}} \) in an AC circuit is given by:
\[ P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
Substituting the values:
\[ P_{\text{avg}} = \left(50\sqrt{2}\right) \cdot \left(10\sqrt{2}\right) \cdot \cos\left(\frac{\pi}{4}\right) \]
We know \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ P_{\text{avg}} = (50\sqrt{2}) \cdot (10\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 1000 \, \text{W} \]

### Step 6: Calculate the wattless current
The wattless current \( I_w \) is given by:
\[ I_w = I_{\text{rms}} \cdot \sin(\phi) \]
Substituting the values:
\[ I_w = (10\sqrt{2}) \cdot \sin\left(\frac{\pi}{4}\right) \]
We know \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ I_w = (10\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 10 \, \text{A} \]

### Final Results
Thus, the average power consumed by the circuit is \( 1000 \, \text{W} \) and the wattless current is \( 10 \, \text{A} \).