When the reverse potential in a semiconductor diode are 10 V and 30 V, thent he corresponding reverse currents are `25muA`. And `50muA` respectively. The reverse resistance of junction diode will be-

Options

A. `40Omega`
B. `4xx10^(5)Omega`
C. `40KOmega`
D. `4xx10^(-5)Omega`

Answer

To find the reverse resistance of the junction diode, we can follow these steps:

### Step 1: Identify the given values
- Reverse potential (V1) = 10 V, Reverse current (I1) = 25 µA = 25 × 10^(-6) A
- Reverse potential (V2) = 30 V, Reverse current (I2) = 50 µA = 50 × 10^(-6) A

### Step 2: Calculate the change in voltage (ΔV)
\[ \Delta V = V2 - V1 = 30 \, \text{V} - 10 \, \text{V} = 20 \, \text{V} \]

### Step 3: Calculate the change in current (ΔI)
\[ \Delta I = I2 - I1 = 50 \, \mu A - 25 \, \mu A = (50 - 25) \times 10^{-6} \, A = 25 \times 10^{-6} \, A \]

### Step 4: Use Ohm's Law to find the reverse resistance (R)
According to Ohm's law, resistance can be calculated using the formula:
\[ R = \frac{\Delta V}{\Delta I} \]
Substituting the values we calculated:
\[ R = \frac{20 \, \text{V}}{25 \times 10^{-6} \, A} \]

### Step 5: Calculate the resistance
\[ R = \frac{20}{25 \times 10^{-6}} = \frac{20}{25} \times 10^{6} = 0.8 \times 10^{6} \, \Omega = 8 \times 10^{5} \, \Omega \]

### Step 6: Conclusion
The reverse resistance of the junction diode is \( 8 \times 10^{5} \, \Omega \).

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