Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:Capacitor behavior in first-order RC networks is a cornerstone of electronics. A common misunderstanding is that capacitor voltage can jump instantly. In reality, the defining property of a capacitor is that current is proportional to the rate of change of voltage, which leads to an exponential change in voltage for step inputs—not an instantaneous jump—when driven through a resistor.
Given Data / Assumptions:
Concept / Approach:In an RC network, i_C = C * dv_C/dt. For a finite series resistance, dv_C/dt is finite for finite current; therefore, v_C cannot change instantaneously. For a step input, the solution is v_C(t) = V_final * (1 − exp(−t/τ)) + V_initial * exp(−t/τ), which is exponential in time. Only an ideal zero-impedance source directly across a capacitor (or infinite current) could enforce an instantaneous change—neither occurs in a standard RC integrator.
Step-by-Step Solution:
Identify that the current into the capacitor is limited by the series resistor.Relate current and voltage rate: i_C = C * dv_C/dt ⇒ finite i_C gives finite dv_C/dt.For a step input, write the first-order solution showing exponential charging/discharging with τ = R * C.Conclude that capacitor voltage changes exponentially, not instantaneously, in an RC integrator.Verification / Alternative check:Oscilloscope measurements of an RC charge curve show the well-known exponential approach to a new level, reaching ~63% in 1τ, ~86% in 2τ, ~95% in 3τ, ~98% in 4τ, and ~99% in 5τ. None of these represent an instantaneous jump.
Why Other Options Are Wrong:
Common Pitfalls:Confusing an observed fast but finite rise with an instantaneous jump; ignoring series resistance or source limitations that necessarily enforce exponential transitions.
Final Answer:Incorrect.
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