A p-n junction diode when forward baiased has a drop of 0.5 V which is assumed to be independent of current. The current in excess of 10 mA through the diode produces a large joule heating which damages (burns) the diode. If we want to use a 15.V battery to forward bias diode a resistor of resistance `axx10^(2)Omega` is to be used in serise with the diode so that the maximum current does not exceed 5mA. what is the value of a?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to determine the value of \( a \) such that the maximum current through the diode does not exceed 5 mA when connected to a 1.5 V battery with a resistor of resistance \( a \times 10^2 \, \Omega \) in series with the diode.

### Step-by-Step Solution:

1. **Identify the total voltage and voltage drop across the diode:**
- The total voltage from the battery is \( V = 1.5 \, \text{V} \).
- The voltage drop across the diode when forward biased is \( V_D = 0.5 \, \text{V} \).

2. **Calculate the voltage across the resistor:**
- The voltage across the resistor \( V_R \) can be calculated using:
\[ V_R = V - V_D = 1.5 \, \text{V} - 0.5 \, \text{V} = 1.0 \, \text{V} \]

3. **Use Ohm's Law to find the resistance needed to limit the current:**
- The maximum current \( I_{max} \) through the diode is given as \( 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \).
- According to Ohm's Law, \( V_R = I \times R \), we can rearrange this to find \( R \):
\[ R = \frac{V_R}{I_{max}} = \frac{1.0 \, \text{V}}{5 \times 10^{-3} \, \text{A}} = 200 \, \Omega \]

4. **Relate the resistance to the given form:**
- The resistance is given as \( R = a \times 10^2 \, \Omega \).
- We found \( R = 200 \, \Omega \), so we can set up the equation:
\[ a \times 10^2 = 200 \]

5. **Solve for \( a \):**
- Dividing both sides by \( 10^2 \):
\[ a = \frac{200}{100} = 2 \]

Thus, the value of \( a \) is \( 2 \).

### Final Answer:
\[ \text{The value of } a \text{ is } 2. \]