Step-down chopper with 1 V drop: average and rms

With an ON-state drop $V_{ch}$, the load voltage is a pulse of height $(V_s - V_{ch})$ lasting $kT$ in every period, so the average and rms values follow directly from that pulse.

Given: $V_s = 230$ V, $R = 10\ \Omega$, $V_{ch} = 1$ V, $k = 0.4$.

 v0   229|_____       _____
         |     |     |     |
       0 |_____|_____|_____|____ t
          kT    T

Average output voltage

$$\begin{aligned} V_a &= \frac{1}{T}\int_0^{kT}(V_s - V_{ch})\,dt = k\,(V_s - V_{ch}) \\ &= 0.4\times(230-1) = 0.4\times229 = 91.6\ \text{V} \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms} &= \left[\frac{1}{T}\int_0^{kT}(V_s - V_{ch})^2dt\right]^{1/2} = \sqrt{k}\,(V_s - V_{ch}) \\ &= \sqrt{0.4}\times229 = 0.63246\times229 = 144.83\ \text{V} \end{aligned}$$

For reference, the load currents are $I_a = 91.6/10 = 9.16$ A and $I_{rms} = 14.48$ A.

Answer: $V_a = 91.6$ V, $V_{rms} = 144.83$ V.