An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?

Explanation

To solve the problem step by step, we will follow the process of finding the RMS current through the capacitor when connected to the given alternating voltage.

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

### Step 2: Find the angular frequency \( \omega \)
From the voltage equation, we can identify:
\[ \omega = 100 \, \text{rad/s} \]

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

### Step 4: Find the peak voltage \( E_0 \)
The peak voltage \( E_0 \) is:
\[ E_0 = 200\sqrt{2} \, \text{V} \]

### Step 5: Calculate the peak current \( I_0 \)
The peak current \( I_0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{E_0}{X_C} \]
Substituting the values:
\[ I_0 = \frac{200\sqrt{2}}{10^4} = \frac{200\sqrt{2}}{10000} = \frac{\sqrt{2}}{50} \]

### Step 6: Calculate the RMS current \( I_{RMS} \)
The RMS current \( I_{RMS} \) is related to the peak current by:
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} \]
Substituting the value of \( I_0 \):
\[ I_{RMS} = \frac{\frac{\sqrt{2}}{50}}{\sqrt{2}} = \frac{1}{50} \, \text{A} = 0.02 \, \text{A} \]

### Step 7: Conclusion
The ammeter reads the RMS value of the current, which we calculated as:
\[ I_{RMS} = 0.02 \, \text{A} \]

### Final Answer
The reading of the ammeter will be:
\[ \text{0.02 A} \]

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