In a silicon diode, the reverse current increases from `10 mu A `to `20 mu A`, when the reverse voltage changes from 2 to 4V.The reverse ac resistance of the diode is

Options

A. `1 xx 10^5 Omega `
B. `3 xx 10^5 Omega `
C. `2 xx 10^5 Omega `
D. `4 xx 10^5 Omega `

Answer

To find the reverse AC resistance of a silicon diode, we can follow these steps:

### Step 1: Identify the change in reverse current (ΔI)
The reverse current increases from \(10 \mu A\) to \(20 \mu A\). Therefore, the change in current (ΔI) can be calculated as follows:
\[ \Delta I = I_{final} - I_{initial} = 20 \mu A - 10 \mu A = 10 \mu A \]
Converting microamperes to amperes:
\[ \Delta I = 10 \times 10^{-6} A \]

### Step 2: Identify the change in reverse voltage (ΔV)
The reverse voltage changes from \(2 V\) to \(4 V\). Thus, the change in voltage (ΔV) is:
\[ \Delta V = V_{final} - V_{initial} = 4 V - 2 V = 2 V \]

### Step 3: Calculate the reverse AC resistance (Rr)
The reverse AC resistance (Rr) can be calculated using the formula:
\[ R_r = \frac{\Delta V}{\Delta I} \]
Substituting the values we found:
\[ R_r = \frac{2 V}{10 \times 10^{-6} A} \]
Calculating this gives:
\[ R_r = \frac{2}{10 \times 10^{-6}} = 2 \times 10^{5} \, \Omega \]

### Conclusion
The reverse AC resistance of the diode is \(2 \times 10^{5} \, \Omega\).

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