An alternating voltage (in volts) given by `V=200sqrt(2)sin(100t)` is connected to `1 mu F` capacitor through an ideal ac ammeter in series. The reading of the ammeter and the average power consumed in the circuit shall be

Options

A. 20 mA, 0
B. 20 mA, 4W
C. `20sqrt(2)mA, 8W`
D. `20sqrt(2)mA, 4sqrt(2)W`

Explanation

To solve the problem, we need to find the reading of the ammeter and the average power consumed in the circuit when an alternating voltage is applied to a capacitor. Let's break it down step by step.

### Step 1: Identify the given values
The alternating voltage is given by:
\[ V = 200\sqrt{2} \sin(100t) \]
The capacitance of the capacitor is:
\[ C = 1 \mu F = 1 \times 10^{-6} F \]

### Step 2: Determine the peak voltage (\( V_0 \)) and angular frequency (\( \omega \))
From the voltage equation, we can identify:
- Peak voltage \( V_0 = 200\sqrt{2} \) volts
- Angular frequency \( \omega = 100 \) rad/s

### Step 3: Calculate the RMS voltage (\( V_{rms} \))
The RMS voltage can be calculated using the formula:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
Substituting the value of \( V_0 \):
\[ V_{rms} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \text{ volts} \]

### Step 4: Calculate the capacitive reactance (\( X_c \))
The capacitive reactance is given by the formula:
\[ X_c = \frac{1}{\omega C} \]
Substituting the values of \( \omega \) and \( C \):
\[ X_c = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \text{ ohms} \]

### Step 5: Calculate the RMS current (\( I_{rms} \))
Using the formula for current in an AC circuit:
\[ I_{rms} = \frac{V_{rms}}{X_c} \]
Substituting the values:
\[ I_{rms} = \frac{200}{10^4} = 20 \times 10^{-3} \text{ A} = 20 \text{ mA} \]

### Step 6: Calculate the average power consumed (\( P \))
The average power consumed in a capacitor circuit is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
In a purely capacitive circuit, the phase difference \( \phi \) is \( 90^\circ \), so:
\[ \cos(90^\circ) = 0 \]
Thus, the average power consumed is:
\[ P = 200 \cdot 20 \times 10^{-3} \cdot 0 = 0 \text{ watts} \]

### Final Answers:
- The reading of the ammeter is \( 20 \text{ mA} \).
- The average power consumed in the circuit is \( 0 \text{ watts} \).