Power factor of an L-R series circuit is 0.6 and that of a C-R series circuit is 0.5. If the element (L. C, and R) of the two circuits are joined in series 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

Answer

To solve the problem step by step, we will analyze the given information about the L-R and C-R circuits, use trigonometric relationships to find the necessary values, and derive the ratio of resistances.

### Step 1: Understand the Power Factor of the L-R Circuit
The power factor (PF) of the L-R circuit is given as 0.6.
- We can express this as:
\[ \cos \phi_1 = 0.6 = \frac{3}{5} \]
- From the power triangle, we know:
\[ \sin \phi_1 = \sqrt{1 - \cos^2 \phi_1} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 2: Find the Relationship Between Resistance and Reactance in the L-R Circuit
Using the definitions of tangent:
- We have:
\[ \tan \phi_1 = \frac{\sin \phi_1}{\cos \phi_1} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]
- In the L-R circuit, \( \tan \phi_1 = \frac{X_L}{R_1} \):
\[ \frac{X_L}{R_1} = \frac{4}{3} \implies R_1 = \frac{3}{4} X_L \quad \text{(Equation 1)} \]

### Step 3: Understand the Power Factor of the C-R Circuit
The power factor of the C-R circuit is given as 0.5.
- We can express this as:
\[ \cos \phi_2 = 0.5 = \frac{1}{2} \]
- From the power triangle, we find:
\[ \sin \phi_2 = \sqrt{1 - \cos^2 \phi_2} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 4: Find the Relationship Between Resistance and Reactance in the C-R Circuit
Using the definitions of tangent:
- We have:
\[ \tan \phi_2 = \frac{\sin \phi_2}{\cos \phi_2} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]
- In the C-R circuit, \( \tan \phi_2 = \frac{X_C}{R_2} \):
\[ \frac{X_C}{R_2} = \sqrt{3} \implies R_2 = \frac{X_C}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 5: Use the Resonance Condition
For the L-C-R circuit, the power factor is 1, which indicates resonance:
- At resonance, \( X_L = X_C \).

### Step 6: Find the Ratio of Resistances
Now we can find the ratio \( \frac{R_1}{R_2} \):
- From Equation 1:
\[ R_1 = \frac{3}{4} X_L \]
- From Equation 2:
\[ R_2 = \frac{X_C}{\sqrt{3}} \]
- Since \( X_L = X_C \) at resonance, we can substitute:
\[ \frac{R_1}{R_2} = \frac{\frac{3}{4} X_L}{\frac{X_L}{\sqrt{3}}} = \frac{3}{4} \cdot \sqrt{3} = \frac{3\sqrt{3}}{4} \]

### Final Answer
The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is:
\[ \frac{R_1}{R_2} = \frac{3\sqrt{3}}{4} \]