Step-down chopper with 2 V drop: average and rms

With an ON-state drop of $V_{ch}$, the load receives $(V_s - V_{ch})$ for $kT$ and zero for the rest of the period.

Given: $V_s = 230$ V, $R = 10\ \Omega$, $V_{ch} = 2$ V, $k = 0.4$.

 v0   228|_____       _____
         |     |     |     |
       0 |_____|_____|_____|____ t
          kT    T

Average output voltage

$$\begin{aligned} V_a &= \frac{1}{T}\int_0^{kT}(V_s - V_{ch})\,dt = k\,(V_s - V_{ch}) \\ &= 0.4\times(230 - 2) = 0.4\times228 = 91.2\ \text{V} \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms} &= \left[\frac{1}{T}\int_0^{kT}(V_s - V_{ch})^2\,dt\right]^{1/2} = \sqrt{k}\,(V_s - V_{ch}) \\ &= \sqrt{0.4}\times228 = 0.63246\times228 = 144.20\ \text{V} \end{aligned}$$

Corresponding load currents: $I_a = 91.2/10 = 9.12$ A, $I_{rms} = 14.42$ A. Chopper efficiency would be $(V_s - V_{ch})/V_s = 228/230 = 99.13\%$.

Answer: $V_a = 91.2$ V, $V_{rms} = 144.20$ V.