The power factor of an L-R series circuit is 0.5 and that of a C-R series circuit is 0.2. If the elements (L, C and R) of the two circuit are joined in series and connected to the same ac source, the power factor of this circuit is found to be 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is

Options

A. 2
B. `sqrt2`
C. `2sqrt2`
D. 4

Answer

To solve the problem, we need to analyze the given information about the power factors of the L-R and C-R circuits and derive the relationship between the resistances in those circuits.

### Step-by-Step Solution:

1. **Understand the Power Factor**:
The power factor (PF) is defined as the cosine of the phase angle (φ) between the voltage and current in an AC circuit. For an L-R circuit, the power factor is given by:
\[ \cos \phi_1 = \frac{R_1}{Z_1} \]
where \(Z_1\) is the impedance of the L-R circuit.

2. **Write the Equation for L-R Circuit**:
Given that the power factor for the L-R circuit is 0.5:
\[ \cos \phi_1 = 0.5 \implies \frac{R_1}{\sqrt{R_1^2 + X_L^2}} = 0.5 \]
Squaring both sides, we get:
\[ R_1^2 = 0.25(R_1^2 + X_L^2) \implies R_1^2 = 0.25R_1^2 + 0.25X_L^2 \]
Rearranging gives:
\[ 0.75R_1^2 = 0.25X_L^2 \implies 3R_1^2 = X_L^2 \quad \text{(Equation 1)} \]

3. **Write the Equation for C-R Circuit**:
For the C-R circuit, the power factor is given as 0.2:
\[ \cos \phi_2 = 0.2 \implies \frac{R_2}{\sqrt{R_2^2 + X_E^2}} = 0.2 \]
Squaring both sides gives:
\[ R_2^2 = 0.04(R_2^2 + X_E^2) \implies R_2^2 = 0.04R_2^2 + 0.04X_E^2 \]
Rearranging gives:
\[ 0.96R_2^2 = 0.04X_E^2 \implies 24R_2^2 = X_E^2 \quad \text{(Equation 2)} \]

4. **Write the Equation for LCR Circuit**:
For the LCR circuit, the power factor is 1, which implies:
\[ \cos \phi_3 = 1 \implies R = \sqrt{R^2 + (X_L - X_E)^2} \]
Since \(\cos \phi_3 = 1\), we have:
\[ R^2 + (X_L - X_E)^2 = R^2 \implies (X_L - X_E)^2 = 0 \implies X_L = X_E \quad \text{(Equation 3)} \]

5. **Combine Equations**:
From Equations 1 and 2, we have:
\[ X_L^2 = 3R_1^2 \quad \text{and} \quad X_E^2 = 24R_2^2 \]
Since \(X_L = X_E\), we can equate:
\[ 3R_1^2 = 24R_2^2 \implies \frac{R_1^2}{R_2^2} = \frac{24}{3} = 8 \]
Taking the square root of both sides gives:
\[ \frac{R_1}{R_2} = 2\sqrt{2} \]

### Final Answer:
The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is:
\[ \frac{R_1}{R_2} = 2\sqrt{2} \]