Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NEET2020PhysicsAlternating CurrentActual

A series L C R circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is π 3 . If instead C is removed from the circuit, the phase difference is again π 3 between current and voltage. The power factor of the circuit is:

Options

  1. A0.5
  2. B1.0
  3. C– 1.0
  4. Dzero

Correct answer

B. 1.0

Step-by-step solution

As phase angle contribution by capacitor and inductor are equal in magnitude, so ( gathered X _ L = X _ c L = 1 C gathered ) It means that the RLC circuit is in resonance condition. So, impedance at resonance, (Z= R^2+ (X_L-X_C )^2 = R^2+ (X_L-X_L )^2 = R^2 =R ) Thus, power factor, ( = R Z = R R =1 )

Practice Alternating Current on Quantrex Academy →

More from Alternating Current

A step down transformer connected to an a.c. mains of 220 V is made to operate at 5.5 V, 44 W lamp. The current in the primary circuit is (Ignore power losses) 2026Which phasor diagram represents LCR circuit at resonance? 2026For an R-L series circuit, the power factor is 3 2 , for R-L frequency f Hz. If the frequency doubles, the new power factor will be 2026In an AC circuit, the current is I = 100 (5t) A. The value of I_ rms is 2026The ratio of power factor of purely resistive circuit to purely reactive circuit is ( 0^ = 1 and 90^ = 0 ) 2026In an LCR series circuit, at resonance, 2026In an AC circuit, E and I are given by E = 150 (150t) V and I = 150 (150t + 3 ) A. The power dissipated in the circuit is (60)^ = 1/2 2026A series LCR circuit is connected across a source E of e.m.f. E=15 (50 t- 3 ) . The current from the supply is I=5 (50 t+ 6 ) . The impedance of the circuit and the phase differenc 2026 Full Alternating Current list All NEET PYQs