What is the capacitive reactance of a capacitor of 5μF at a frequency (1) 50 Hz and (2) 20KHZ?

Answer

To find the capacitive reactance (Xc) of a capacitor at different frequencies, 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 \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- Frequency (1) \( f_1 = 50 \, Hz \)
- Frequency (2) \( f_2 = 20 \, kHz = 20 \times 10^{3} \, Hz \)

### Step-by-Step Solution:

**Step 1: Calculate Xc for 50 Hz**
1. Substitute the values into the formula:
\[ X_{c1} = \frac{1}{2 \pi (50) (5 \times 10^{-6})} \]
2. Calculate \( 2 \pi \):
\[ 2 \pi \approx 6.28 \]
3. Now calculate:
\[ X_{c1} = \frac{1}{6.28 \times 50 \times 5 \times 10^{-6}} \]
4. Calculate \( 6.28 \times 50 \):
\[ 6.28 \times 50 = 314 \]
5. Now calculate:
\[ X_{c1} = \frac{1}{314 \times 5 \times 10^{-6}} = \frac{1}{1.57 \times 10^{-3}} \approx 636.94 \, \Omega \]

**Step 2: Calculate Xc for 20 kHz**
1. Substitute the values into the formula:
\[ X_{c2} = \frac{1}{2 \pi (20 \times 10^{3}) (5 \times 10^{-6})} \]
2. Again, calculate \( 2 \pi \):
\[ 2 \pi \approx 6.28 \]
3. Now calculate:
\[ X_{c2} = \frac{1}{6.28 \times (20 \times 10^{3}) \times (5 \times 10^{-6})} \]
4. Calculate \( 6.28 \times 20 \times 10^{3} \):
\[ 6.28 \times 20 \times 10^{3} = 125600 \]
5. Now calculate:
\[ X_{c2} = \frac{1}{125600 \times 5 \times 10^{-6}} = \frac{1}{0.628} \approx 1.59 \, \Omega \]

### Final Answers:
- Capacitive reactance at 50 Hz: \( X_{c1} \approx 636.94 \, \Omega \)
- Capacitive reactance at 20 kHz: \( X_{c2} \approx 1.59 \, \Omega \)