Explanation:

In a centre tapped full wave rectifier circuit, the peak voltage is the maximum voltage that appears across the load resistor when the input AC voltage is at its peak. The peak voltage is given by the formula:

Peak Voltage = (2 * Vm) - Vr

Where:
- Vm is the peak voltage of the input AC voltage
- Vr is the voltage drop across the diode(s)

In this case, the peak voltage is given as 5V.

Calculating the Voltage Drop Across the Diode:

The voltage drop across a diode is typically given as the diode's forward voltage drop, also known as the diode cut-in voltage. In this case, the diode cut-in voltage is given as 0.7V.

Substituting the Values:

Using the formula for peak voltage, we can substitute the given values to find the voltage drop across the diode:

5V = (2 * Vm) - 0.7V

Simplifying the equation:

5V = 2 * Vm - 0.7V
5V + 0.7V = 2 * Vm
5.7V = 2 * Vm

Calculating the Peak Inverse Voltage:

The peak inverse voltage is the maximum voltage that appears across the diode(s) when the input AC voltage is at its peak. It is equal to twice the peak voltage.

Therefore, the peak inverse voltage is given by:

Peak Inverse Voltage = 2 * Peak Voltage

Substituting the value of the peak voltage:

Peak Inverse Voltage = 2 * 5V
Peak Inverse Voltage = 10V

So, the correct answer is option 'B': 9.3V.

Summary:

- The peak voltage is the maximum voltage that appears across the load resistor in a centre tapped full wave rectifier circuit.
- The voltage drop across a diode is typically given as the diode's forward voltage drop.
- The peak inverse voltage is the maximum voltage that appears across the diode(s) and is equal to twice the peak voltage.
- In this question, the peak voltage is given as 5V and the diode cut-in voltage is 0.7V.
- Substituting these values, we find that the peak inverse voltage is 9.3V.