A resistor of resistance R, capacitor of capacitance C and inductor of inductance L are connected in parallel to AC power source of voltage `epsi_0` sin `omegat`. The maximum current through the resistance is half of the maximum current through the power source. Then value of R is

Options

A. `(sqrt(3))/(|omegaC - 1/(omega L)|)`
B. `sqrt3 |1/(omegaC) - omegaL|`
C. `sqrt5 |1/(omegaC) - omegaL|`
D. None of these

Answer

To solve the problem, we need to analyze the circuit with a resistor (R), capacitor (C), and inductor (L) connected in parallel to an AC power source. The maximum current through the resistor is given to be half of the maximum current through the power source.

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- We have a resistor (R), capacitor (C), and inductor (L) connected in parallel to an AC source with voltage \( \epsilon_0 \sin(\omega t) \).

2. **Identify Maximum Currents**:
- The maximum current through the resistor \( I_R \) is given by:
\[ I_R = \frac{\epsilon_0}{R} \]
- The maximum current through the capacitor \( I_C \) is given by:
\[ I_C = \epsilon_0 \cdot \omega C \]
- The maximum current through the inductor \( I_L \) is given by:
\[ I_L = \frac{\epsilon_0}{\omega L} \]

3. **Relationship Between Currents**:
- According to the problem, the maximum current through the resistor is half of the maximum current through the power source:
\[ I_R = \frac{1}{2} I_{max} \]
- The total maximum current \( I_{max} \) in the circuit can be expressed using the Pythagorean theorem since the currents through R, C, and L are at right angles:
\[ I_{max} = \sqrt{I_R^2 + (I_C - I_L)^2} \]

4. **Substituting Known Values**:
- Substitute \( I_R \), \( I_C \), and \( I_L \) into the equation:
\[ I_{max} = \sqrt{\left(\frac{\epsilon_0}{R}\right)^2 + \left(\epsilon_0 \cdot \omega C - \frac{\epsilon_0}{\omega L}\right)^2} \]

5. **Setting Up the Equation**:
- Since \( I_R = \frac{1}{2} I_{max} \), we can write:
\[ \frac{\epsilon_0}{R} = \frac{1}{2} \sqrt{\left(\frac{\epsilon_0}{R}\right)^2 + \left(\epsilon_0 \cdot \omega C - \frac{\epsilon_0}{\omega L}\right)^2} \]

6. **Squaring Both Sides**:
- Square both sides to eliminate the square root:
\[ \left(\frac{\epsilon_0}{R}\right)^2 = \frac{1}{4} \left(\left(\frac{\epsilon_0}{R}\right)^2 + \left(\epsilon_0 \cdot \omega C - \frac{\epsilon_0}{\omega L}\right)^2\right) \]

7. **Simplifying the Equation**:
- Multiply through by 4 to clear the fraction:
\[ 4 \left(\frac{\epsilon_0}{R}\right)^2 = \left(\frac{\epsilon_0}{R}\right)^2 + \left(\epsilon_0 \cdot \omega C - \frac{\epsilon_0}{\omega L}\right)^2 \]
- Rearranging gives:
\[ 3 \left(\frac{\epsilon_0}{R}\right)^2 = \left(\epsilon_0 \cdot \omega C - \frac{\epsilon_0}{\omega L}\right)^2 \]

8. **Dividing by \( \epsilon_0^2 \)**:
- Dividing both sides by \( \epsilon_0^2 \):
\[ 3 \frac{1}{R^2} = \left(\omega C - \frac{1}{\omega L}\right)^2 \]

9. **Finding R**:
- Taking the square root gives:
\[ \frac{1}{R} = \sqrt{3} \left(\omega C - \frac{1}{\omega L}\right) \]
- Therefore, the value of \( R \) is:
\[ R = \frac{1}{\sqrt{3} \left(\omega C - \frac{1}{\omega L}\right)} \]

### Final Answer:
\[ R = \frac{1}{\sqrt{3} \left(\omega C - \frac{1}{\omega L}\right)} \]