An alternating voltage given as, `V=100sqrt(2)sin100t` V is applied to a capacitor of `1muF`. The current reading of the ammeter will be equal to ................ mA.

Options

A. 20
B. 10
C. 40
D. 80

Explanation

To find the current reading of the ammeter when an alternating voltage is applied to a capacitor, we can follow these steps:

### Step 1: Identify the given parameters
The alternating voltage is given as:
\[ V(t) = 100\sqrt{2} \sin(100t) \, \text{V} \]
From this, we can identify:
- The peak voltage \( V_0 = 100\sqrt{2} \, \text{V} \)
- The angular frequency \( \omega = 100 \, \text{rad/s} \)

The capacitance \( C \) is given as:
\[ C = 1 \, \mu F = 1 \times 10^{-6} \, F \]

### Step 2: Calculate the RMS voltage
The RMS (Root Mean Square) voltage \( V_{rms} \) can be calculated using the formula:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
Substituting the value of \( V_0 \):
\[ V_{rms} = \frac{100\sqrt{2}}{\sqrt{2}} = 100 \, \text{V} \]

### Step 3: Calculate the capacitive reactance \( X_c \)
The capacitive reactance \( X_c \) is given by:
\[ X_c = \frac{1}{\omega C} \]
Substituting the values of \( \omega \) and \( C \):
\[ X_c = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{0.0001} = 10000 \, \Omega \]

### Step 4: Calculate the current \( I \)
The current \( I \) through the capacitor can be calculated using the formula:
\[ I = \frac{V_{rms}}{X_c} \]
Substituting the values of \( V_{rms} \) and \( X_c \):
\[ I = \frac{100}{10000} = 0.01 \, A \]

### Step 5: Convert the current to milliamperes
To express the current in milliamperes (mA), we multiply by 1000:
\[ I = 0.01 \, A \times 1000 = 10 \, mA \]

### Final Answer
The current reading of the ammeter will be equal to:
\[ \boxed{10 \, mA} \]