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 varepsilon₀sinω t. The maximum current through the resistance is half of the maximum current through the power source. Then value of R is:

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

Hint:

In parallel AC circuits: I=√(IR²+(IC-IL)²) Always compare vector currents, not algebraic sums.

Answer 1

The Correct Option is B

Step 1: In a parallel RLC circuit, the current through the resistor: IR=(V₀)/(R)
Step 2: Net source current: I=√(IR²+(IC-IL)²)
Step 3: Given IR=(I)/(2): I²=4IR² ⟹ (IC-IL)²=3IR²
Step 4: IC-IL=V₀(ω C-(1)/(ω L))
Step 5: (V₀)/(R)=\frac1√(3)V₀|ω C-(1)/(ω L)| ⟹ R=√(3)|(1)/(ω C)-ω L|