In a full wave junction diode rectifier, the input a.c. has r.m.s. value of 20 V. The transformer used is a step up transformer having primary and secondary turns ratio 1: 2. Calculate the d.c. voltade in the recitified output.

Options

A. 12 V
B. 24 V
C. 36 V
D. 42 V

Explanation

To solve the problem of finding the d.c. voltage in the rectified output of a full wave junction diode rectifier with the given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the RMS Value**:
The RMS value of the input AC voltage is given as \( V_{rms} = 20 \, V \).

2. **Calculate the Peak Voltage**:
The peak voltage \( V_{peak} \) can be calculated using the formula:
\[ V_{peak} = V_{rms} \times \sqrt{2} \]
Substituting the given RMS value:
\[ V_{peak} = 20 \times \sqrt{2} \approx 20 \times 1.414 \approx 28.28 \, V \]

3. **Determine the Transformer Output Voltage**:
Since the transformer is a step-up transformer with a turns ratio of 1:2, the output voltage \( V' \) from the transformer can be calculated as:
\[ V' = 2 \times V_{peak} \]
Substituting the peak voltage we calculated:
\[ V' = 2 \times 28.28 \approx 56.56 \, V \]

4. **Calculate the DC Voltage Output**:
For a full-wave rectifier, the average (DC) output voltage \( V_{dc} \) can be approximated using the formula:
\[ V_{dc} = \frac{2 \times V'}{\pi} \]
Substituting the output voltage from the transformer:
\[ V_{dc} = \frac{2 \times 56.56}{\pi} \approx \frac{113.12}{3.14} \approx 36 \, V \]

5. **Final Answer**:
The d.c. voltage in the rectified output is approximately \( 36 \, V \).