In an RL differentiator, when the input pulse transitions from its low level to its high level, which statement is correct?

Introduction / Context:
Inductors resist changes in current, which is central to how RL differentiators produce sharp voltage spikes at edges. Recognizing this physical property helps predict transient behavior in pulse circuits.

Given Data / Assumptions:

• Series RL with output typically taken across the resistor (differentiator behavior for small tau).
• Input undergoes a rising step (pulse goes from low to high).

Concept / Approach:
Inductor voltage is v_L = L * di/dt. A finite inductance cannot instantaneously change current without requiring infinite voltage. Therefore, at the instant of a step, current through the inductor remains (momentarily) at its prior value, and the required voltage appears to support a rapid change in current thereafter.

Step-by-Step Solution:

Verification / Alternative check:
Ideal inductor i_L is continuous; measured waveforms show resistor voltage spikes ±V proportional to di/dt. Simulations or scope captures confirm that i_L ramps rather than jumps at the transition.

Why Other Options Are Wrong:

• Prevents sudden change in voltage: That is the property of a capacitor, not an inductor.
• Instantly reaches 63%: 63% relates to one time constant, not an instantaneous effect.
• Voltage across inductor is zero: At an edge, v_L is typically largest, not zero.

Common Pitfalls:

• Confusing capacitor and inductor edge properties.
• Assuming the 63% rule applies at t = 0 rather than at t = tau.

Final Answer:
The inductor prevents a sudden change in current