A `P-N` junction diode can withstand currents up to `10 mA`. Under forward bias, The diode has a potential drop of `0.5 V` across it which is assumed to be independent of current. The maximum voltage of the battery used to forward bias the diode when a resistance of `200 Omega` is connected in series with it is

Options

A. `2.5 V`
B. `2.6 V`
C. `2.7 V`
D. `2.8 V`

Explanation

To find the maximum voltage of the battery used to forward bias the diode when a resistance of 200 ohms is connected in series with it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Maximum current (I) = 10 mA = \(10 \times 10^{-3}\) A
- Voltage drop across the diode (V_d) = 0.5 V
- Resistance (R) = 200 Ω

2. **Use Ohm's Law**:
According to Ohm's law, the voltage across the resistor can be calculated using the formula:
\[ V_R = I \times R \]
Substituting the values:
\[ V_R = (10 \times 10^{-3}) \times 200 = 2 \text{ V} \]

3. **Calculate Total Voltage (E)**:
The total voltage supplied by the battery (E) is the sum of the voltage drop across the diode (V_d) and the voltage across the resistor (V_R):
\[ E = V_d + V_R \]
Substituting the known values:
\[ E = 0.5 \text{ V} + 2 \text{ V} = 2.5 \text{ V} \]

4. **Conclusion**:
The maximum voltage of the battery used to forward bias the diode is:
\[ E = 2.5 \text{ V} \]

### Final Answer:
The maximum voltage of the battery is **2.5 V**.

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