In a full wave rectifiers in which input voltage is represented by `V=V_(M)sin omegat`, then peak inversion voltage of non-conducting diode will be

Options

A. `-V_(M)`
B. `V_(M)//2`
C. `2V_(M)`
D. 0

Explanation

To solve the problem, we need to determine the peak inversion voltage of a non-conducting diode in a full wave rectifier circuit where the input voltage is given by \( V = V_m \sin(\omega t) \).

### Step-by-Step Solution:

1. **Understand the Input Voltage**:
The input voltage is given as \( V = V_m \sin(\omega t) \), where \( V_m \) is the peak voltage of the input AC signal.

2. **Identify the Function of the Full Wave Rectifier**:
A full wave rectifier converts both halves of the AC waveform into DC. During the positive half cycle of the input AC voltage, one diode conducts while the other diode is reverse-biased (non-conducting). During the negative half cycle, the roles of the diodes are reversed.

3. **Determine the Peak Inversion Voltage**:
The peak inversion voltage (PIV) is the maximum reverse voltage that a diode can withstand without conducting. In a full wave rectifier, the PIV for a non-conducting diode is equal to the sum of the peak voltage of the input AC signal from both diodes.

4. **Calculate the Peak Inversion Voltage**:
For a full wave rectifier, the peak inversion voltage \( V_p \) is given by:
\[ V_p = 2 \times V_m \]
This is because during the negative half cycle, the non-conducting diode experiences the full peak voltage \( V_m \) from the other diode plus the peak voltage from the input AC source.

5. **Final Answer**:
Therefore, the peak inversion voltage of the non-conducting diode is:
\[ V_p = 2 V_m \]

### Conclusion:
The peak inversion voltage of the non-conducting diode in a full wave rectifier with input voltage \( V = V_m \sin(\omega t) \) is \( 2 V_m \).