A series LCR circuit is connected across an AC power supply of 220 V. The frequency of source is variable . The vaues of L and C are 0.320 H and 0.32 `muF ` respectively. Calculate the value of frequency to be applied such that voltage across R is maximum .

Answer

To solve the problem, we need to determine the frequency at which the voltage across the resistor \( R \) in a series LCR circuit is maximized. This occurs when the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Inductance \( L = 0.320 \, H \)
- Capacitance \( C = 0.32 \, \mu F = 0.32 \times 10^{-6} \, F \)
- Voltage \( V = 220 \, V \) (not needed for frequency calculation)

2. **Understand the Condition for Maximum Voltage Across \( R \):**
- The voltage across the resistor \( R \) is maximum when the circuit is at resonance, which occurs when the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).
- Mathematically, this is expressed as:
\[ X_L = X_C \]

3. **Write the Formulas for Reactances:**
- Inductive reactance:
\[ X_L = \omega L = 2 \pi f L \]
- Capacitive reactance:
\[ X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} \]

4. **Set the Reactances Equal:**
\[ 2 \pi f L = \frac{1}{2 \pi f C} \]

5. **Rearrange to Find Frequency:**
- Multiply both sides by \( 2 \pi f C \):
\[ (2 \pi f)^2 L C = 1 \]
- Therefore,
\[ (2 \pi f)^2 = \frac{1}{LC} \]
- Taking the square root:
\[ 2 \pi f = \frac{1}{\sqrt{LC}} \]
- Thus,
\[ f = \frac{1}{2 \pi \sqrt{LC}} \]

6. **Substitute the Values of \( L \) and \( C \):**
\[ f = \frac{1}{2 \pi \sqrt{0.320 \times 0.32 \times 10^{-6}}} \]

7. **Calculate \( LC \):**
- First, calculate \( LC \):
\[ LC = 0.320 \times 0.32 \times 10^{-6} = 0.1024 \times 10^{-6} = 1.024 \times 10^{-7} \]

8. **Calculate \( \sqrt{LC} \):**
\[ \sqrt{LC} = \sqrt{1.024 \times 10^{-7}} \approx 0.00032 \]

9. **Calculate Frequency \( f \):**
\[ f = \frac{1}{2 \pi \times 0.00032} \approx \frac{1}{0.0002} \approx 500 \, Hz \]

### Final Answer:
The frequency to be applied such that the voltage across \( R \) is maximum is approximately \( 500 \, Hz \).

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