A simple capacitor filter fed by a full-wave rectifier develops 14.5 V dc at 8.5% ripple factor. What is the output ripple voltage (rms)?
$$ \% r=\frac{V_{r(\mathrm{~ms})}}{V_{\mathrm{dc}}} \times 100 \% $$
$$ V_{r(\mathrm{~ms})}=\frac{r \times V_{\mathrm{dc}}}{100} \ldots \ldots (1)$$
Substitute \(8.5\) for \(r\), and \(14.5 \mathrm{~V}\) for \(V_{\mathrm{d}}\) in equation (1).
$$ \begin{aligned} V_{r(\mathrm{~ms})} &=\frac{(8.5)(14.5 \mathrm{~V})}{100} \\ &=\frac{123.25}{100} \\ &=1.23 \mathrm{~V} \end{aligned} $$