A resistance is connected to a capacitor in AC are the phase differece is `(pi)/(4)` between current and voltage. Whe the same resistance is connected to an inductor, phase difference becomes `tan^(-1)(2)`. Power factor of the circuit when both capacitor and inductor are connect to the resistance will be

Options

A. 1
B. `(1)/(sqrt2)`
C. `(1)/(sqrt3)`
D. `(1)/(2)`

Answer

To solve the problem step by step, we will analyze the given information regarding the phase differences when a resistance is connected to a capacitor and an inductor, and then find the power factor when both are connected.

### Step 1: Understand the Phase Difference with the Capacitor
When the resistance \( R \) is connected to a capacitor, the phase difference \( \phi_C \) is given as \( \frac{\pi}{4} \) (or 45 degrees). The power factor \( \cos(\phi_C) \) can be expressed as:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{R}{Z_C} \]
where \( Z_C \) is the impedance of the circuit with the capacitor.

### Step 2: Calculate the Capacitive Reactance
From the above equation, we can express the impedance \( Z_C \):
\[ Z_C = \frac{R}{\cos\left(\frac{\pi}{4}\right)} = R \sqrt{2} \]
Now, we know that:
\[ Z_C = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance. Since \( Z_C = R \sqrt{2} \), we can set up the equation:
\[ R \sqrt{2} = \sqrt{R^2 + X_C^2} \]
Squaring both sides:
\[ 2R^2 = R^2 + X_C^2 \]
This simplifies to:
\[ X_C^2 = R^2 \implies X_C = R \]

### Step 3: Understand the Phase Difference with the Inductor
When the same resistance \( R \) is connected to an inductor, the phase difference \( \phi_L \) is given by:
\[ \tan(\phi_L) = 2 \]
This implies:
\[ \tan(\phi_L) = \frac{X_L}{R} = 2 \implies X_L = 2R \]

### Step 4: Calculate the Impedance with Both Capacitor and Inductor
When both the capacitor and inductor are connected to the resistance, the total impedance \( Z \) can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values of \( X_L \) and \( X_C \):
\[ Z = \sqrt{R^2 + (2R - R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

### Step 5: Calculate the Power Factor
The power factor \( \text{PF} \) is given by:
\[ \text{PF} = \frac{R}{Z} \]
Substituting the value of \( Z \):
\[ \text{PF} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the power factor of the circuit when both the capacitor and inductor are connected to the resistance is:
\[ \text{Power Factor} = \frac{1}{\sqrt{2}} \]