A `200 muF` capacitor in series with a `100 Omega` resistance is connected to a 240 V, 50 Hz supply. What is the maximum current in the circuit ?

Options

A. `1.4 A`
B. `3.4 A`
C. `4.4A`
D. `2.4 A`

Answer

To find the maximum current in the given RC circuit, we can follow these steps:

### Step 1: Identify the given values
- Capacitance, \( C = 200 \mu F = 200 \times 10^{-6} F \)
- Resistance, \( R = 100 \Omega \)
- RMS Voltage, \( V_{rms} = 240 V \)
- Frequency, \( f = 50 Hz \)

### Step 2: Calculate the maximum voltage (\( V_0 \))
The relationship between the RMS voltage and the maximum voltage is given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
To find \( V_0 \):
\[ V_0 = V_{rms} \times \sqrt{2} = 240 \times \sqrt{2} \approx 240 \times 1.414 = 339.41 V \]

### Step 3: Calculate the capacitive reactance (\( X_C \))
The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2 \pi f \).
Calculating \( \omega \):
\[ \omega = 2 \pi \times 50 \approx 314.16 \, rad/s \]
Now substituting into the formula for \( X_C \):
\[ X_C = \frac{1}{314.16 \times 200 \times 10^{-6}} \approx \frac{1}{0.06283} \approx 15.92 \, \Omega \]

### Step 4: Calculate the impedance (\( Z \))
The total impedance in an RC circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
Substituting the values:
\[ Z = \sqrt{100^2 + 15.92^2} = \sqrt{10000 + 253.4464} = \sqrt{10253.4464} \approx 101.26 \, \Omega \]

### Step 5: Calculate the maximum current (\( I_0 \))
The maximum current can be calculated using:
\[ I_0 = \frac{V_0}{Z} \]
Substituting the values:
\[ I_0 = \frac{339.41}{101.26} \approx 3.35 \, A \]

### Final Answer
The maximum current in the circuit is approximately \( 3.35 \, A \).

---