RL filter identification (series network): In a simple series R–L circuit driven by a sinusoidal source, if the output is measured across the resistor (voltage across R), does the network behave as a low-pass filter that passes low frequencies and attenuates high frequencies?

Introduction / Context:
First-order RL circuits are the magnetic dual of RC filters. Whether they act as low-pass or high-pass depends entirely on where the output is taken. In practice, the series RL measured across the resistor is a classic one-pole low-pass used for current sensing and simple anti-alias filtering.

Given Data / Assumptions:

• Single loop: sinusoidal source feeding a series resistor R and inductor L.
• Output node is the voltage across R (not across L).
• Linear, time-invariant operation with ideal components.

Concept / Approach:
The inductor’s reactance is XL = 2 * pi * f * L. As frequency increases, XL increases and the series current decreases. Since v_R = i * R, the output magnitude falls with frequency. Conversely, at low frequency XL is small, current is larger, and v_R approaches the input, which is the hallmark of a low-pass response. The cutoff frequency (−3 dB) occurs at f_c = R / (2 * pi * L).

Step-by-Step Solution:

Verification / Alternative check:
Bode magnitude shows −20 dB/decade roll-off above f_c, and the phase of V_R lags less than 90 degrees, approaching 0 degrees at low frequency—typical of a one-pole low-pass across a resistive element in a series RL path.

Why Other Options Are Wrong:
High-pass behavior occurs when the output is taken across L, not R. Limiting L to zero removes filtering. The response is broadband (a full passband and stopband), not a single-frequency effect.

Common Pitfalls:
Confusing where the output is measured; swapping to the inductor output flips the function to high-pass. Forgetting that the cutoff depends on R/L, not on R or L alone.

Final Answer:
Applies (low-pass across the series resistor)