In a reverse biased diode, when the applied voltage changes by `1V`, the current is found to change by `0.5muA`. The reversebiase resistance of the diode is

Options

A. ` 2 xx 10^(5) Omega`
B. ` 2 xx 10^(6) Omega`
C. `200 Omega`
D. `2 Omega`

Answer

To find the reverse bias resistance of the diode, we can use Ohm's law, which states that:

\[ V = I \cdot R \]

Where:
- \( V \) is the voltage across the diode,
- \( I \) is the current flowing through the diode,
- \( R \) is the resistance.

### Step-by-Step Solution:

1. **Identify the Change in Voltage and Current**:
- The change in voltage (\( \Delta V \)) is given as \( 1 \, \text{V} \).
- The change in current (\( \Delta I \)) is given as \( 0.5 \, \mu A = 0.5 \times 10^{-6} \, A \).

2. **Use Ohm's Law to Find Resistance**:
- Rearranging Ohm's law to solve for resistance gives us:
\[ R = \frac{V}{I} \]
- In this case, we will use the changes in voltage and current:
\[ R = \frac{\Delta V}{\Delta I} \]

3. **Substitute the Values**:
- Substitute \( \Delta V = 1 \, V \) and \( \Delta I = 0.5 \times 10^{-6} \, A \):
\[ R = \frac{1 \, V}{0.5 \times 10^{-6} \, A} \]

4. **Calculate the Resistance**:
- Performing the calculation:
\[ R = \frac{1}{0.5 \times 10^{-6}} = \frac{1}{0.5} \times 10^{6} = 2 \times 10^{6} \, \Omega \]

5. **Conclusion**:
- The reverse bias resistance of the diode is \( 2 \times 10^{6} \, \Omega \).

### Final Answer:
The reverse bias resistance of the diode is \( 2 \times 10^{6} \, \Omega \).

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