Frequency change in a <em>series</em> RL circuit (R and L fixed): If the operating frequency decreases, which of the following statements is correct for the circuit's parameters and responses?

Introduction / Context:In a series RL circuit, several quantities depend on the operating frequency. The most direct dependency is the inductive reactance X_L = 2 * pi * f * L. Understanding how X_L and related responses change with frequency is vital for filter design and power systems where frequency variations can shift current, voltage drops, and phase angle.

Given Data / Assumptions:

• Series RL network with fixed R and L.
• Steady-state sinusoidal operation.
• We analyze qualitative change when frequency f decreases.

Concept / Approach:The most fundamental effect is on inductive reactance: X_L = 2 * pi * f * L. If f decreases, X_L decreases proportionally. Consequences include a smaller impedance magnitude |Z| = sqrt(R^2 + X_L^2), larger current I = V_S / |Z| (for fixed V_S), and a smaller phase angle theta = arctan(X_L / R). Among the choices, the statement that is unambiguously and directly true for any values is that X_L decreases.

Step-by-Step Solution:

Verification / Alternative check:Pick L = 100 mH. At 50 Hz, X_L ≈ 31.4 Ω; at 25 Hz, X_L ≈ 15.7 Ω. The halved frequency halves X_L, confirming the rule.

Why Other Options Are Wrong:

• I_T decreases: opposite; current increases as |Z| falls (for fixed V_S).
• I_R increases / I_L increases: in a series circuit I_R = I_L = I_T, but stating only one branch increases is incomplete and ambiguous; the fundamental certain statement concerns X_L directly.
• The phase angle increases: it actually decreases because X_L/R becomes smaller.

Common Pitfalls:Confusing series and parallel cases; assuming only the inductive branch current changes; overlooking that all series branch currents are identical and follow I = V_S / |Z|.

Final Answer:Inductive reactance decreases