A full-wave rectified signal of 18 V peak is fed into a capacitor filter. What is the voltage regulation of the filter if the output is 17 V dc at full load?
The no load voltage of the full-wave rectifier is equal to the maximum voltage of the signal.
$$ \begin{aligned} V_{N L} &=V_{m} \\ &=18 \mathrm{~V} \end{aligned} $$
Write the expression for percentage voltage regulation for filter.
$$ \% \mathrm{~V} \cdot \mathrm{R} .=\frac{V_{\mathrm{NL}}-V_{\mathrm{FL}}}{V_{\text {s. }}} \times 100 \% \ldots \ldots \text { (1) } $$
Substitute \(18 \mathrm{~V}\) for \(V_{\mathrm{NL}}\), and \(17 \mathrm{~V}\) for \(V_{\mathrm{FL}}\) to calculate \(\% \mathrm{~V} . \mathrm{R} .\) in equation \((1)\)
$$ \begin{aligned} \% \text { V.R. } &=\frac{18 \mathrm{~V}-17 \mathrm{~V}}{17 \mathrm{~V}} \times 100 \% \\ &=\frac{1}{17} \times 100 \% \\ &=5.88 \% \end{aligned} $$
Therefore, the percentage voltage regulation for filter is \(5.88 \%\).