Parallel RL frequency effect: When the frequency of the applied AC source is decreased, how does the total impedance of a parallel RL circuit change?

Introduction / Context:
Understanding how frequency affects impedance is key for filter design and AC circuit analysis. In a parallel RL circuit, the resistive branch admittance is constant with frequency, while the inductive branch admittance varies with frequency.

Given Data / Assumptions:

• Parallel RL topology.
• Inductive reactance XL = 2 * pi * f * L.
• Admittance of the inductor YL = 1 / (j XL) = -j / XL.

Concept / Approach:
When frequency decreases, XL decreases. In parallel circuits we add admittances: Y_total = G + jB, where G = 1/R and B = -1/XL. If XL decreases, |B| increases in magnitude, making |Y_total| larger. Since Z_total = 1 / |Y_total|, the total impedance decreases.

Step-by-Step Solution:
XL = 2 * pi * f * LDecrease f ⇒ decrease XLYL = 1 / (j XL) ⇒ |YL| increases as XL decreasesY_total increases ⇒ Z_total = 1 / |Y_total| decreases

Verification / Alternative check:
Numerical example: Let R = 100 Ω, L = 0.1 H. At f1 = 100 Hz, XL ≈ 62.8 Ω; at f2 = 50 Hz, XL ≈ 31.4 Ω. The inductive branch admittance doubles in magnitude, so the net admittance increases and net impedance drops, confirming the trend.

Why Other Options Are Wrong:

• Increases: Opposite of parallel behavior driven by inductive branch admittance.
• Remains constant / not a factor: Frequency directly affects XL, so impedance must change.

Common Pitfalls:
Applying series-circuit intuition to parallel circuits. In series RL, lowering frequency reduces XL and thus reduces total impedance; in parallel RL, the effect flips because we combine admittances, not impedances.

Final Answer:
decreases