This question was previously asked in

ESE Electronics 2014 Paper 1: Official Paper

Option 4 : 8.1 V

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

**Concept:**

CIRCUIT |
Number of Diodes |
Average DC Voltage (Vdc) |
RMS Current (Irms) |
Peak Inverse Voltage (PIV) |

Half-Wave Rectifier |
1 |
\(\frac{{{V_m}}}{\pi }\) |
\(\frac{{{I_{m\;}}}}{2}\) |
\({V_m}\) |

Center-Tap Full Wave Rectifier |
2 |
\(\frac{{2{V_m}}}{\pi }\) |
\(\frac{{{I_m}}}{{\sqrt 2 }}\) |
\(2{V_m}\) |

Bridge-Type Full Wave Rectifier |
4 |
\(\frac{{2{V_m}}}{\pi }\) |
\(\frac{{{I_m}}}{{\sqrt 2 }}\) |
\({V_m}\) |

__Calculation:__

Given that, input voltage (V_{s}) = 9 V

V_{m} = √2 × V_{s} = 9√2 V

Output voltage \({V_0} = \frac{{2 \times 9\sqrt 2 }}{\pi } = 8.1\;V\)