A capacitor of capacitance `10muF` is connected to an AC source and an AC Ammeter. If the source voltage varies as `V=50sqrt(2)sin100t`, the reading of the ammeter is

Options

A. 50 mA
B. 70.7 mA
C. 5.0 mA
D. 7.07 mA

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Capacitance \( C = 10 \mu F = 10 \times 10^{-6} F \)
- Voltage \( V(t) = 50\sqrt{2} \sin(100t) \)

### Step 2: Determine the angular frequency \( \omega \)
From the voltage equation, we can see that:
- \( \omega = 100 \, \text{rad/s} \)

### Step 3: Calculate the capacitive reactance \( X_c \)
The formula for capacitive reactance is given by:
\[ X_c = \frac{1}{\omega C} \]
Substituting the values:
\[ X_c = \frac{1}{100 \times 10 \times 10^{-6}} = \frac{1}{10^{-3}} = 1000 \, \Omega \]

### Step 4: Calculate the RMS voltage \( V_{rms} \)
The RMS voltage for a sinusoidal voltage is given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
Where \( V_0 = 50\sqrt{2} \):
\[ V_{rms} = \frac{50\sqrt{2}}{\sqrt{2}} = 50 \, V \]

### Step 5: Calculate the RMS current \( I_{rms} \)
Using Ohm's law for AC circuits, the RMS current is given by:
\[ I_{rms} = \frac{V_{rms}}{X_c} \]
Substituting the values:
\[ I_{rms} = \frac{50}{1000} = 0.05 \, A = 50 \, mA \]

### Step 6: Convert to microamperes
Since \( 1 \, A = 10^6 \, \mu A \):
\[ I_{rms} = 50 \, mA = 50 \times 10^3 \, \mu A = 50 \, \mu A \]

### Final Answer
The reading of the ammeter is \( 50 \, \mu A \).

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