Question

A resistor of resistance R, capacitor of capacitance C and inductor of inductance L are connected in parallel to an AC power source of voltage \( \varepsilon_0 \sin \omega t \). The maximum current through the resistance is half of the maximum current through the power source. Then value of R is:

Answer:
\( R = \sqrt{3}\left| \dfrac{1}{\omega C} - \omega L \right| \)

• A. \(\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right|\)
• B. \(\sqrt{3}\left|\frac{1}{\omega C}+\omega L\right|\)
• C. \(\sqrt{5}\left|\frac{1}{\omega C}-\omega L\right|\)
• D. None of these

Hint:

In parallel AC circuits, currents add vectorially, not algebraically.

Answer 1

The Correct Option is A

Step 1: Current through resistor:
\( I_R = \dfrac{\varepsilon_0}{R} \)
Step 2: Total current:
\( I = \varepsilon_0 \sqrt{\left(\dfrac{1}{R}\right)^2 + \left(\omega C - \dfrac{1}{\omega L}\right)^2} \)
Step 3: Given \( I_R = \dfrac{I}{2} \):
\( \dfrac{1}{R} = \dfrac{1}{2} \sqrt{\left(\dfrac{1}{R}\right)^2 + \left(\omega C - \dfrac{1}{\omega L}\right)^2} \)
Step 4: Solving:
\( R = \sqrt{3}\left| \dfrac{1}{\omega C} - \omega L \right| \)