Problem

# A full-wave rectified signal of 18 V peak is fed into a capacitor filter. What is the volt...

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?

#### Step-by-Step Solution

Solution 1

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 $$5.88 \%$$.