NEET 2018 · PhysicsAverage powerPrevious Year Question
An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is :
Answer: (A) 0.51 W. Answer: (A) 0.51 W Solution: X_L = ωL = 340 × 20×10⁻³ = 6.8 Ω X_C = 1/(ωC) = 1/(340 × 50×10⁻⁶) ≈ 58.8 Ω Z = √[(X_L − X_C)² + R²] = √[(6.8 − 58.8)² + 40²] ≈ 65.6 Ω I_rms = V_rms/Z = (10/√2)/65.6 Average power P = I_rms²·R = (100/(65.6²·2))·40 = 2000/65.6² ≈ 0.51 W.
- A.0.51 W✓
- B.0.67 W
- C.0.76 W
- D.0.89 W
Correct Answer
(A) 0.51 W
Solution & Explanation
Answer: (A) 0.51 W Solution: X_L = ωL = 340 × 20×10⁻³ = 6.8 Ω X_C = 1/(ωC) = 1/(340 × 50×10⁻⁶) ≈ 58.8 Ω Z = √[(X_L − X_C)² + R²] = √[(6.8 − 58.8)² + 40²] ≈ 65.6 Ω I_rms = V_rms/Z = (10/√2)/65.6 Average power P = I_rms²·R = (100/(65.6²·2))·40 = 2000/65.6² ≈ 0.51 W.
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