A light emitting diode `(LED)` has a voltage drop of `2V` across it and passes a current of `10 mA`. When it operates with a `6V` battery through a limiting resistor `R`. The value of `R` is

Options

A. `40kOmega`
B. `4 kOmega`
C. `200Omega`
D. `400Omega`

Explanation

To solve the problem, we need to determine the value of the limiting resistor \( R \) when a light emitting diode (LED) operates with a 6V battery. The LED has a voltage drop of 2V and passes a current of 10mA.

### Step-by-Step Solution:

1. **Identify the given values:**
- Voltage drop across the LED, \( V_{LED} = 2V \)
- Current through the LED, \( I = 10mA = 10 \times 10^{-3} A = 0.01 A \)
- Voltage of the battery, \( V_{battery} = 6V \)

2. **Calculate the voltage across the resistor \( R \):**
- The voltage across the resistor can be found using the formula:
\[ V_R = V_{battery} - V_{LED} \]
- Substituting the values:
\[ V_R = 6V - 2V = 4V \]

3. **Use Ohm's Law to find the resistance \( R \):**
- Ohm's Law states that:
\[ V = I \cdot R \]
- Rearranging this gives:
\[ R = \frac{V_R}{I} \]
- Substituting the values we have:
\[ R = \frac{4V}{0.01 A} \]

4. **Calculate the resistance:**
- Performing the calculation:
\[ R = \frac{4}{0.01} = 400 \, \Omega \]

5. **Conclusion:**
- The value of the limiting resistor \( R \) is \( 400 \, \Omega \).

### Final Answer:
The value of \( R \) is \( 400 \, \Omega \).

---