Q189 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
A full-wave rectifier provides a DC output voltage of $20\ \text{V}$. If the root mean square (RMS) value of the unwanted AC component is measured as $2\ \text{V}$, the ripple factor in percentage is
- A$2\%$
- B$5\%$
- Correct answerC$10\%$
- D$20\%$
Solution
Given the DC output voltage $V_{dc} = 20\ \text{V}$ and the RMS value of the AC component $V_{ac} = 2\ \text{V}$.
The ripple factor $\gamma$ is defined as the ratio of the RMS value of the AC component to the DC component of the output voltage:
$\gamma = \dfrac{V_{ac}}{V_{dc}}$
Substituting the given values:
$\gamma = \dfrac{2}{20} = 0.1$
Expressing the ripple factor in percentage:
$\gamma = 0.1 \times 100\% = 10\%$
Answer: $10\%$
The ripple factor $\gamma$ is defined as the ratio of the RMS value of the AC component to the DC component of the output voltage:
$\gamma = \dfrac{V_{ac}}{V_{dc}}$
Substituting the given values:
$\gamma = \dfrac{2}{20} = 0.1$
Expressing the ripple factor in percentage:
$\gamma = 0.1 \times 100\% = 10\%$
Answer: $10\%$
Hence, the correct answer is option: 'C'.