What is the capacitance reactance of a `5mu F` capacitor when it is part of a circuit whose frequency is (i) 50 Hz (ii) `10^(6)` Hz ?

Answer

To find the capacitive reactance (Xc) of a capacitor, we can use the formula:

\[ X_c = \frac{1}{2 \pi f C} \]

where:
- \( X_c \) is the capacitive reactance in ohms (Ω),
- \( f \) is the frequency in hertz (Hz),
- \( C \) is the capacitance in farads (F).

Given:
- Capacitance \( C = 5 \mu F = 5 \times 10^{-6} F \)

We will calculate the capacitive reactance for two different frequencies: (i) 50 Hz and (ii) \( 10^6 \) Hz.

### Step 1: Calculate Xc for f = 50 Hz

1. Substitute the values into the formula:
\[ X_c = \frac{1}{2 \pi (50) (5 \times 10^{-6})} \]

2. Calculate \( 2 \pi (50) \):
\[ 2 \pi (50) \approx 314.16 \]

3. Now, substitute this value back into the equation:
\[ X_c = \frac{1}{314.16 \times 5 \times 10^{-6}} \]

4. Calculate \( 314.16 \times 5 \times 10^{-6} \):
\[ 314.16 \times 5 \times 10^{-6} \approx 1.5708 \times 10^{-3} \]

5. Finally, calculate \( X_c \):
\[ X_c \approx \frac{1}{1.5708 \times 10^{-3}} \approx 636.94 \, \Omega \]

### Step 2: Calculate Xc for f = \( 10^6 \) Hz

1. Substitute the values into the formula:
\[ X_c = \frac{1}{2 \pi (10^6) (5 \times 10^{-6})} \]

2. Calculate \( 2 \pi (10^6) \):
\[ 2 \pi (10^6) \approx 6.2832 \times 10^6 \]

3. Now, substitute this value back into the equation:
\[ X_c = \frac{1}{6.2832 \times 10^6 \times 5 \times 10^{-6}} \]

4. Calculate \( 6.2832 \times 10^6 \times 5 \times 10^{-6} \):
\[ 6.2832 \times 5 \approx 31.416 \]

5. Finally, calculate \( X_c \):
\[ X_c \approx \frac{1}{31.416} \approx 0.0318 \, \Omega \]

### Summary of Results
- For \( f = 50 \, Hz \), \( X_c \approx 636.94 \, \Omega \)
- For \( f = 10^6 \, Hz \), \( X_c \approx 0.0318 \, \Omega \)