A DC chopper applies a constant input voltage V to a resistive load for a fraction a (duty cycle) of each period and 0 for the remainder. What is the RMS value of the output voltage across the load?

Difficulty: Easy

Correct Answer: V * sqrt(a)

Explanation:


Introduction / Context:
Pulse-width-modulated (PWM) choppers feed a resistive load with a rectangular waveform: V during on-time and 0 during off-time. While the average output is a*V, the RMS value is different and is important for heating and true power calculations in resistive loads.



Given Data / Assumptions:

  • Input DC = V (constant).
  • Duty cycle = a (0 < a ≤ 1).
  • Load is purely resistive; ripple current follows voltage instantly.


Concept / Approach:

RMS value is defined as sqrt( (1/T) * ∫ v^2(t) dt ). For a two-level waveform (V for aT, 0 for (1−a)T), square the levels, integrate over the period, and take the square root. This gives a simple closed form that differs from the average aV.



Step-by-Step Solution:

v(t) = V for duration aT; v(t) = 0 for (1−a)T.v^2(t) = V^2 over aT; 0 over (1−a)T.V_rms = sqrt( (1/T)*(V^2*aT) ) = sqrt( V^2 * a ) = V * sqrt(a).Thus RMS and average are different: V_rms = V*sqrt(a), V_avg = aV.


Verification / Alternative check:

Check limits: a → 1 gives V_rms → V (continuous DC), a → 0 gives V_rms → 0, both consistent.



Why Other Options Are Wrong:

  • aV: average value, not RMS.
  • a^2V, V/√a, V/a: incorrect scaling; fail limit checks at a → 0 or a → 1.


Common Pitfalls:

  • Confusing RMS with average (both relevant but distinct).
  • Forgetting to square before integrating for RMS.


Final Answer:

V * sqrt(a)

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