A resistance of `10Omega` is joined in series with an inductance of 0.5H. What capacitance should be put in series with the combination to obtain the maximum current ? What will be this maximum current ? What will be the potential difference across the resistance, inductance and capacitance ? The current is being supplied by 200 V, 50Hz AC mains.

Answer

To solve the problem step by step, we will follow the concepts of RLC circuits, resonance conditions, and the relationships between voltage, current, and impedance.

### Step 1: Identify the Given Values
- Resistance, \( R = 10 \, \Omega \)
- Inductance, \( L = 0.5 \, H \)
- Supply Voltage, \( V_{rms} = 200 \, V \)
- Frequency, \( f = 50 \, Hz \)

### Step 2: Calculate the Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 50 = 100 \pi \, rad/s \]

### Step 3: Determine the Inductive Reactance
The inductive reactance \( X_L \) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100 \pi \times 0.5 = 50 \pi \, \Omega \]

### Step 4: Condition for Maximum Current
For maximum current in an RLC circuit, the inductive reactance \( X_L \) must equal the capacitive reactance \( X_C \):
\[ X_L = X_C \]
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
Setting \( X_L = X_C \):
\[ 50 \pi = \frac{1}{100 \pi C} \]

### Step 5: Solve for Capacitance \( C \)
Rearranging the equation to find \( C \):
\[ C = \frac{1}{50 \pi \times 100 \pi} = \frac{1}{5000 \pi^2} \]
Calculating the value:
\[ C \approx \frac{1}{5000 \times 9.87} \approx 0.0000203 \, F = 20.3 \, \mu F \]

### Step 6: Calculate Maximum Current \( I_{max} \)
The maximum current \( I_{max} \) can be calculated using Ohm's law:
\[ I_{max} = \frac{V_{rms}}{R} \]
Substituting the values:
\[ I_{max} = \frac{200}{10} = 20 \, A \]

### Step 7: Calculate Voltage Across Each Component
1. **Voltage across the Resistance \( V_R \)**:
\[ V_R = I_{max} \times R = 20 \times 10 = 200 \, V \]

2. **Voltage across the Inductor \( V_L \)**:
\[ V_L = I_{max} \times X_L = 20 \times 50\pi \approx 20 \times 157.08 \approx 3141.6 \, V \]

3. **Voltage across the Capacitor \( V_C \)**:
Since \( V_L = V_C \) at resonance,
\[ V_C = V_L \approx 3141.6 \, V \]

### Summary of Results
- Capacitance \( C \approx 20.3 \, \mu F \)
- Maximum Current \( I_{max} = 20 \, A \)
- Voltage across Resistance \( V_R = 200 \, V \)
- Voltage across Inductor \( V_L \approx 3141.6 \, V \)
- Voltage across Capacitor \( V_C \approx 3141.6 \, V \)