In a series LCR circuit, voltage applied is `V=3sin(314t+(pi)/(6))` and current from the supply is `l=2sin(315t+(pi)/(3))`. Which of the following is correct ?

Options

A. Impendance of circuit is `1.5Omega`
B. Reactance of circuit is `(4)/(3)Omega`
C. Resistance of circuit is `(3sqrt3)/(4)Omega`
D. Wattless component of current is `(1)/(sqrt2)`

Answer

To solve the problem, we need to analyze the given voltage and current equations in the LCR circuit and determine which statements about the circuit are correct.

### Given:
- Voltage: \( V = 3 \sin(314t + \frac{\pi}{6}) \)
- Current: \( I = 2 \sin(315t + \frac{\pi}{3}) \)

### Step 1: Identify the parameters from the voltage and current equations.

From the voltage equation:
- Amplitude of voltage, \( V_0 = 3 \)
- Angular frequency of voltage, \( \omega_v = 314 \)
- Phase angle of voltage, \( \phi_v = \frac{\pi}{6} \)

From the current equation:
- Amplitude of current, \( I_0 = 2 \)
- Angular frequency of current, \( \omega_i = 315 \)
- Phase angle of current, \( \phi_i = \frac{\pi}{3} \)

### Step 2: Calculate the impedance of the circuit.

In a series LCR circuit, the impedance \( Z \) can be calculated using the formula:
\[ Z = \frac{V_0}{I_0} \]
Substituting the values we have:
\[ Z = \frac{3}{2} = 1.5 \, \text{ohm} \]

### Step 3: Determine the reactance of the circuit.

The reactance \( X \) in a series LCR circuit can be calculated using:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]
However, we do not have the values of \( L \) and \( C \) provided in the problem, so we cannot calculate the reactance directly. Thus, we cannot confirm the reactance statement.

### Step 4: Calculate the average power in the circuit.

The average power \( P \) in the circuit can be calculated using:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\delta \phi) \]
Where \( \delta \phi = \phi_i - \phi_v \):
\[ \delta \phi = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]

Calculating \( V_{rms} \) and \( I_{rms} \):
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{3}{\sqrt{2}}, \quad I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Now substituting into the power formula:
\[ P = \left(\frac{3}{\sqrt{2}}\right) \cdot \left(\sqrt{2}\right) \cdot \cos\left(\frac{\pi}{6}\right) \]
\[ P = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

### Step 5: Calculate the resistance of the circuit.

Using the relationship between power and resistance:
\[ P = I_{rms}^2 \cdot R \]
Substituting \( I_{rms} = \sqrt{2} \):
\[ \frac{3\sqrt{3}}{2} = (\sqrt{2})^2 \cdot R \]
\[ \frac{3\sqrt{3}}{2} = 2R \implies R = \frac{3\sqrt{3}}{4} \, \text{ohm} \]

### Step 6: Calculate the wattless component of current.

The wattless component of current is given by:
\[ I_{wattless} = I_{rms} \cdot \cos(\phi_i) \]
Substituting \( I_{rms} = \sqrt{2} \) and \( \phi_i = \frac{\pi}{3} \):
\[ I_{wattless} = \sqrt{2} \cdot \cos\left(\frac{\pi}{3}\right) = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

### Summary of Results:
1. Impedance \( Z = 1.5 \, \text{ohm} \) (Correct)
2. Reactance cannot be determined (Incorrect)
3. Resistance \( R = \frac{3\sqrt{3}}{4} \, \text{ohm} \) (Correct)
4. Wattless component of current \( = \frac{1}{\sqrt{2}} \) (Correct)

### Final Answer:
- The correct statements are:
- Impedance of the circuit is \( 1.5 \, \text{ohm} \).
- Resistance of the circuit is \( \frac{3\sqrt{3}}{4} \, \text{ohm} \).
- Wattless component of current is \( \frac{1}{\sqrt{2}} \).