A diode has potential drop of 0.5 volts in forward bias. The maximumcurrent that can be flow through diode is 10mA.Then find the resistance connected in series with diode so that set up can be connected to a battery of 1.5 volts:

Options

A. `100ohm`
B. `50ohm`
C. `25ohm`
D. `10ohm`

Answer

To solve the problem step by step, let's analyze the situation with the given data:

1. **Identify the given values:**
- Forward potential drop across the diode (V_diode) = 0.5 V
- Maximum current through the diode (I_max) = 10 mA = 10 × 10^(-3) A
- Total battery voltage (V_battery) = 1.5 V

2. **Determine the voltage across the resistor:**
The voltage across the resistor (V_resistor) can be calculated by subtracting the diode's forward voltage drop from the total battery voltage:
\[ V_{resistor} = V_{battery} - V_{diode} = 1.5 V - 0.5 V = 1.0 V \]

3. **Use Ohm's Law to find the resistance:**
According to Ohm's Law, the relationship between voltage (V), current (I), and resistance (R) is given by:
\[ V = I \times R \]
Rearranging this formula to find R gives:
\[ R = \frac{V}{I} \]
Substituting the values we have:
\[ R = \frac{V_{resistor}}{I_{max}} = \frac{1.0 V}{10 \times 10^{-3} A} \]

4. **Calculate the resistance:**
\[ R = \frac{1.0}{10 \times 10^{-3}} = \frac{1.0}{0.01} = 100 \, \Omega \]

Thus, the resistance that should be connected in series with the diode is **100 Ohms**.