NEETPhysicsAlternating Current
An AC voltage V = 250 (100 t) V is applied to a series LCR circuit containing a resistor of R = 60 , , an inductor of L = 1.2 H, and a capacitor of C = 250 , F. Match the quantities in List-I with their corresponding values in List-II. List-I List-II (A) Inductive reactance (I) 40 , (B) Capacitive reactance (II) 2.5 A (C) Impedance (III) 100 , (D) Current amplitude (IV) 120 , Choose the correct answer from the option
Options
- A(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- B(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
- C(A)-(IV), (B)-(I), (C)-(I), (D)-(III)
- D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Correct answer
A. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
Step-by-step solution
From the given voltage equation V = 250 (100 t) V, the peak voltage is V_m = 250 V and the angular frequency is = 100 rad/s. The inductive reactance is X_L = L = 100 1.2 = 120 , . This matches (A) with (IV). The capacitive reactance is X_C = 1 C = 1 100 250 10⁻⁶ = 1 25000 10⁻⁶ = 1 0.025 = 40 , . This matches (B) with (I). The impedance of the series LCR circuit is Z = R^2 + (X_L - X_C)^2 = 60^2 + (120 - 40)^2 = 60^2 + 80^2 = 100 , . This matches (C) with (III). The current amplitude is I_m = V_m Z = 250 100 = 2.5 A