In the AC circuit shown, an unknown box is in series with 10 Ω resistance. The box power factor is 0.5 and the complete circuit has leading power factor √3/2. Find the effective resistance R of the box (Ω). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

In the AC circuit shown, an unknown box is in series with 10 Ω resistance. The box power factor is 0.5 and the complete circuit has leading power factor √3/2. Find the effective resistance R of the box (Ω). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Let unknown box have impedance R − jX (capacitive for leading current). Its power factor 0.5 gives R/√(R² + X²) = 1/2, so X = √3R. Step 2 — Total impedance Z = (10 + R) − jX has leading power factor √3/2, hence X/(10+R) = 1/√3. Step 3 — Substitute √3R/(10+R) = 1/√3 ⇒ 3R = 10 + R. Step 4 — R = 5 Ω.