A series LCR circuit is connected top an AC power source of 220 V , 50 Hz. The values of L and R are 0.41 H and `80 Omega` respectively. What is the value of C if current and voltage are in phase ?

Answer

To solve the problem, we need to find the value of capacitance \( C \) in a series LCR circuit where the current and voltage are in phase. This condition implies that the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Voltage \( V = 220 \, V \)
- Frequency \( f = 50 \, Hz \)
- Inductance \( L = 0.41 \, H \)
- Resistance \( R = 80 \, \Omega \)

2. **Understand the condition for current and voltage to be in phase:**
- For current and voltage to be in phase in an LCR circuit, the inductive reactance \( X_L \) must equal the capacitive reactance \( X_C \).
- Mathematically, this is expressed as:
\[ X_L = X_C \]

3. **Calculate the inductive reactance \( X_L \):**
- The formula for inductive reactance is:
\[ X_L = \omega L \]
- Where \( \omega = 2 \pi f \).
- Substituting the values:
\[ \omega = 2 \pi \times 50 = 100 \pi \, rad/s \]
- Now calculate \( X_L \):
\[ X_L = (100 \pi) \times 0.41 = 41 \pi \, \Omega \]

4. **Set \( X_L \) equal to \( X_C \):**
- The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
- Setting \( X_L \) equal to \( X_C \):
\[ 41 \pi = \frac{1}{\omega C} \]

5. **Rearranging to find \( C \):**
- Substitute \( \omega = 100 \pi \):
\[ 41 \pi = \frac{1}{100 \pi C} \]
- Rearranging gives:
\[ C = \frac{1}{41 \pi \times 100 \pi} \]
- Simplifying:
\[ C = \frac{1}{4100 \pi^2} \]

6. **Calculate the numerical value of \( C \):**
- Using \( \pi \approx 3.14 \):
\[ C \approx \frac{1}{4100 \times (3.14)^2} \approx \frac{1}{4100 \times 9.8596} \approx \frac{1}{40400} \approx 2.475 \times 10^{-5} \, F \]
- Thus, the value of \( C \) is approximately:
\[ C \approx 2.4 \times 10^{-5} \, F \]

### Final Answer:
The value of capacitance \( C \) is approximately \( 2.4 \times 10^{-5} \, F \).