Capacitor response to non-sinusoidal excitation: When a square-wave voltage is applied to a capacitor, how are the resulting voltage and current waveforms affected?

Introduction / Context:Time-domain behavior of capacitors is foundational for filtering and signal shaping. A capacitor resists sudden voltage changes, so applying a square wave reveals its differentiating effect on current and smoothing effect on voltage.

Given Data / Assumptions:

• An ideal capacitor driven by an ideal square-wave voltage source.
• No significant series resistance except any minimal source impedance.
• Observation of resulting voltage across the capacitor and current through it.

Concept / Approach:For a capacitor, i = C * dv/dt. A square wave has nearly instantaneous edges (large dv/dt at transitions), creating sharp current spikes (peaked current). Between transitions, dv/dt ≈ 0, so current falls toward zero while the capacitor voltage exponentially approaches the new level if resistance is present.

Step-by-Step Solution:Apply i = C * dv/dt to a square-wave input.At rising and falling edges, dv/dt is high → large current peaks.During constant-voltage intervals, dv/dt ≈ 0 → current tends to zero.The capacitor voltage cannot jump; it transitions smoothly (rounded edges).

Verification / Alternative check:Viewing on an oscilloscope: current pulses at edges; voltage across the capacitor shows rounded, exponential-like transitions rather than perfect squares.

Why Other Options Are Wrong:(a) Reverses roles of V and I; (c) both rounded is incorrect because current spikes; (d) both peaked is incorrect because capacitor voltage cannot peak instantaneously like current.

Common Pitfalls:Forgetting i = C * dv/dt, or assuming the capacitor voltage can change abruptly, which violates capacitor behavior.

Final Answer:peak the current and round off the voltage waveforms.