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 solve the problem, we need to find the current reading of the ammeter 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 as:
\[ V = 100\sqrt{2} \sin(100t) \, \text{V} \]
From this, we can identify:
- \( V_0 = 100\sqrt{2} \) (the peak voltage)
- \( \omega = 100 \) (the angular frequency)

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

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

### Step 3: Calculate the capacitive reactance
The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values of \( \omega \) and \( C \):
\[ X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{100 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \, \Omega = 10,000 \, \Omega \]

### Step 4: Calculate the current
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 \( X_C \):
\[ I = \frac{100}{10,000} = 0.01 \, A = 10 \, mA \]

### Final Answer
The current reading of the ammeter will be equal to **10 mA**.

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