An alternating voltage given by `V=300sqrt2sin (50t)` (in volts) is connected across a `1muF` capacitor through an AC ammeter. The reading of the ammeter will be

Options

A. 10mA
B. 40mA
C. 100mA
D. 15mA

Answer

To find the reading of the AC ammeter when an alternating voltage is connected across a capacitor, we can follow these steps:

### Step 1: Identify the given parameters
The given alternating voltage is:
\[ V = 300\sqrt{2} \sin(50t) \]
From this, we can identify:
- \( V_0 = 300\sqrt{2} \) volts (the peak voltage)
- \( \omega = 50 \) rad/s (the angular frequency)

### Step 2: Calculate the RMS voltage
The RMS (Root Mean Square) voltage \( V_{\text{rms}} \) can be calculated using the formula:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]
Substituting the value of \( V_0 \):
\[ V_{\text{rms}} = \frac{300\sqrt{2}}{\sqrt{2}} = 300 \text{ volts} \]

### Step 3: Calculate the capacitive reactance \( X_C \)
The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Where:
- \( C = 1 \mu F = 1 \times 10^{-6} F \)
Substituting the values:
\[ X_C = \frac{1}{50 \times 1 \times 10^{-6}} = \frac{1}{50 \times 10^{-6}} = \frac{1}{5 \times 10^{-5}} = 20,000 \text{ ohms} \]

### Step 4: Calculate the RMS current \( I_{\text{rms}} \)
Using Ohm's law for AC circuits, the RMS current \( I_{\text{rms}} \) can be calculated as:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} \]
Substituting the values:
\[ I_{\text{rms}} = \frac{300}{20000} = 0.015 \text{ A} \]

### Step 5: Convert the current to milliampere
To convert the current from amperes to milliamperes:
\[ I_{\text{rms}} = 0.015 \text{ A} = 15 \text{ mA} \]

### Conclusion
The reading of the ammeter will be:
\[ \text{Ammeter reading} = 15 \text{ mA} \]

### Final Answer
The correct option is **15 milliampere**.

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