In power-supply filtering applications, where should an inductor be placed relative to the load, and does it aid or oppose rapid changes in current through the load?

Introduction / Context:
Inductors are often used in power supplies as smoothing elements. Their fundamental behavior is to resist sudden changes in current, which makes them useful in reducing ripple current reaching the load after rectification. This question explores correct placement and qualitative behavior of an inductor used as a filter choke.

Given Data / Assumptions:

• DC output with superimposed ripple from a rectifier.
• Inductor used as a choke in a filter network (e.g., L filter or LC filter).
• Load expects relatively steady current.

Concept / Approach:
For a series choke filter, the inductor is placed in series with the load. Because vL = L * di/dt, the inductor develops a voltage opposing rapid changes in current. High-frequency ripple tries to change current quickly, but the inductor resists, thereby smoothing current. Capacitors, conversely, are commonly placed in shunt to ground to bypass ripple voltage.

Step-by-Step Solution:
Place L in series: current to the load must pass through L.Ripple attempts rapid current change; inductor generates opposing voltage.Result: reduced ripple current through the load, improved smoothing.Often paired with shunt capacitors to form LC or π filters for better attenuation.

Verification / Alternative check:
Frequency-domain view: inductor impedance XL = 2 * π * f * L increases with frequency, so higher-frequency ripple components see greater impedance and are attenuated when L is in series with the load.

Why Other Options Are Wrong:

• Parallel placements described here are capacitor-like roles, not inductive choke roles.
• “Aids current changes”: The opposite of an inductor’s behavior; it opposes di/dt.
• Shorting ripple across the rectifier is unsafe and incorrect for an inductor.

Common Pitfalls:

• Confusing capacitor (shunt) and inductor (series) filter positions.
• Assuming the inductor reduces voltage ripple directly; it primarily smooths current which, with capacitors, yields lower voltage ripple.

Final Answer:
placed in series with the load and opposes any current changes