NEET 2021 · PhysicsPhasor voltagesPrevious Year Question
An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figures. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2. The impedance of the circuit is :
Answer: (B) 5 Ω. Answer: (B) 5 Ω Solution: Net reactive voltage = V_L − V_C = 40 − 10 = 30 V.
- A.4 Ω
- B.5 Ω✓
- C.4√2 Ω
- D.5/√2 Ω
Correct Answer
(B) 5 Ω
Solution & Explanation
Answer: (B) 5 Ω Solution: Net reactive voltage = V_L − V_C = 40 − 10 = 30 V. Source voltage V = √((V_L − V_C)² + V_R²) = √(30² + 40²) = √(900 + 1600) = √2500 = 50 V. Current amplitude given as 10√2, so I_rms = 10√2/√2 = 10 A. Impedance Z = V_rms/I_rms = 50/10 = 5 Ω.
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