When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is :

Options

A. `300 Omega`
B. `200 Omega`
C. `50 Omega`
D. `100 Omega`

Explanation

To solve the problem, we need to find the minimum resistance that should be connected in series with the diode to ensure that the current does not exceed the safe limit of 10 mA when a 1.5 V battery is used.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Voltage drop across the diode, \( V_d = 0.5 \, \text{V} \)
- Maximum current through the diode, \( I_{max} = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} \)
- EMF of the battery, \( V_{battery} = 1.5 \, \text{V} \)

2. **Calculate the Voltage across the Resistor:**
- The total voltage provided by the battery is \( 1.5 \, \text{V} \).
- The voltage drop across the diode is \( 0.5 \, \text{V} \).
- Therefore, the voltage across the resistor \( V_R \) can be calculated as:
\[ V_R = V_{battery} - V_d = 1.5 \, \text{V} - 0.5 \, \text{V} = 1.0 \, \text{V} \]

3. **Use Ohm's Law to Find the Resistance:**
- According to Ohm's Law, the voltage across the resistor is related to the current through it and its resistance by the formula:
\[ V_R = I \cdot R \]
- Rearranging this gives:
\[ R = \frac{V_R}{I} \]
- Substituting the values we have:
\[ R = \frac{1.0 \, \text{V}}{10 \times 10^{-3} \, \text{A}} = \frac{1.0}{0.01} = 100 \, \Omega \]

4. **Conclusion:**
- The minimum resistance that should be connected in series with the diode to ensure that the current does not exceed 10 mA is \( R = 100 \, \Omega \).

### Final Answer:
The minimum resistance to be connected in series with the diode is **100 ohms**.