AIIMS 2006 Physics Diodes MCQ Question

To solve the problem, we need to determine the value of the limiting resistor $R$ used in a circuit with a light-emitting diode (LED). The LED has a voltage drop of 2 volts and is operating with a 6-volt battery while passing a current of 10 mA.

Step-by-step Calculation

1. Identify the Given Values:

   • Voltage drop across the LED, $V_{LED} = 2 \, \text{V}$
   • Total supply voltage, $V_{supply} = 6 \, \text{V}$
   • Current through the LED, $I = 10 \, \text{mA} = 0.01 \, \text{A}$

2. Using Ohm's Law: The total voltage in the circuit is applied across the LED and the resistor $R$. According to Ohm's Law, the voltage across the resistor can be calculated as follows:

   $$V_R = V_{supply} - V_{LED}$$

   Substituting the values:

   $$V_R = 6 \, \text{V} - 2 \, \text{V} = 4 \, \text{V}$$

3. Calculating the Resistance: Now, we can use Ohm's Law, which states that voltage ($V$) is equal to current ($I$) times resistance ($R$):

   $$V_R = I \cdot R$$

   Rearranging this formula to solve for $R$:

   $$R = \frac{V_R}{I}$$

   Substituting the values we have:

   $$R = \frac{4 \, \text{V}}{0.01 \, \text{A}} = 400 \, \Omega$$

Conclusion

Thus, the value of the limiting resistor $R$ is $400 \, \Omega$.

The correct answer is D) 400 Ω.

Clarification of Other Options

Let’s briefly explain why the other options are incorrect:

• A) 40 kΩ: This value would limit the current much below 10 mA, leading to insufficient current for the LED to emit light.

• B) 4 kΩ: This would allow a current of $I = \frac{4 \, \text{V}}{4,000 \, \Omega} = 0.001 \, \text{A} = 1 \, \text{mA}$, which is also too low for typical LED operation.

• C) 200 Ω: This would allow a current of $I = \frac{4 \, \text{V}}{200 \, \Omega} = 0.02 \, \text{A} = 20 \, \text{mA}$, which is above the rated current of 10 mA and could damage the LED.

Summary

For an LED with a voltage drop of 2 V and a current of 10 mA connected to a 6 V battery, the limiting resistor must be $400 \, \Omega$. This ensures the LED operates correctly without exceeding its current rating, making option D the correct choice.