A circuit has a resistance of `50 ohms` and an inductance of `3/pi` henry.It is connected in series with a condenser of `40/pi muF` and `AC` supply voltage of `200 V` and `50` cycles/sec.Calculate (i)the impedance of the circuit.
(ii)the p.d. across inductor coil and condenser.
(iii)Power factor

Answer

To solve the given problem step-by-step, we will calculate the impedance of the circuit, the potential difference across the inductor and capacitor, and the power factor.

### Given Data:
- Resistance, \( R = 50 \, \Omega \)
- Inductance, \( L = \frac{3}{\pi} \, \text{H} \)
- Capacitance, \( C = \frac{40}{\pi} \, \mu\text{F} = \frac{40 \times 10^{-6}}{\pi} \, \text{F} \)
- Frequency, \( f = 50 \, \text{Hz} \)
- Supply Voltage, \( V = 200 \, \text{V} \)

### Step 1: Calculate Angular Frequency (\( \omega \))
\[ \omega = 2\pi f = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

**Hint:** Remember that the angular frequency is calculated by multiplying \( 2\pi \) with the frequency in Hz.

### Step 2: Calculate Inductive Reactance (\( X_L \))
\[ X_L = \omega L = (100\pi) \left(\frac{3}{\pi}\right) = 300 \, \Omega \]

**Hint:** Inductive reactance is given by \( X_L = \omega L \), where \( \omega \) is the angular frequency and \( L \) is the inductance.

### Step 3: Calculate Capacitive Reactance (\( X_C \))
\[ X_C = \frac{1}{\omega C} = \frac{1}{100\pi \left(\frac{40 \times 10^{-6}}{\pi}\right)} = \frac{1}{4 \times 10^{-4}} = 2500 \, \Omega \]

**Hint:** Capacitive reactance is calculated using \( X_C = \frac{1}{\omega C} \).

### Step 4: Calculate Total Impedance (\( Z \))
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
\[ Z = \sqrt{50^2 + (300 - 250)^2} = \sqrt{2500 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \, \Omega \]

**Hint:** The total impedance in an RLC series circuit is found using the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

### Step 5: Calculate the Current (\( I \))
\[ I = \frac{V}{Z} = \frac{200}{50\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{A} \]

**Hint:** The current can be calculated using Ohm's law for AC circuits, \( I = \frac{V}{Z} \).

### Step 6: Calculate Potential Difference Across Inductor (\( V_L \)) and Capacitor (\( V_C \))
\[ V_L = I \cdot X_L = (2\sqrt{2}) \cdot 300 = 600\sqrt{2} \, V \]
\[ V_C = I \cdot X_C = (2\sqrt{2}) \cdot 2500 = 5000\sqrt{2} \, V \]

**Hint:** The potential difference across the inductor and capacitor can be calculated using \( V = I \cdot X \) where \( X \) is the reactance.

### Step 7: Calculate Power Factor (\( \text{pf} \))
\[ \text{pf} = \cos(\phi) = \frac{R}{Z} = \frac{50}{50\sqrt{2}} = \frac{1}{\sqrt{2}} \]

**Hint:** The power factor is the cosine of the phase angle, which can be found using \( \text{pf} = \frac{R}{Z} \).

### Summary of Results:
1. **Impedance (\( Z \))**: \( 50\sqrt{2} \, \Omega \)
2. **Potential Difference Across Inductor (\( V_L \))**: \( 600\sqrt{2} \, V \)
3. **Potential Difference Across Capacitor (\( V_C \))**: \( 5000\sqrt{2} \, V \)
4. **Power Factor**: \( \frac{1}{\sqrt{2}} \)