An LED rated 2 V, 10 mA is connected to a 6 V DC source. What series resistance range ensures safe operation of the LED?
Electronics and Communication Engineering
Electronic Devices and Circuits
Difficulty: Easy
Choose an option
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A0 to 200 Ω
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B200 - 400 Ω
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C200 Ω and above
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D400 Ω and above
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E100 Ω to 150 Ω
Answer
Correct Answer: 400 Ω and above
Explanation
Introduction / Context:For LEDs, a current-limiting resistor is essential to keep the forward current within the rated value. Without it, excessive current can immediately damage the junction.
Given Data / Assumptions:
- Supply voltage Vs = 6 V DC.
- LED forward voltage Vf ≈ 2 V at 10 mA.
- Rated forward current If = 10 mA.
Concept / Approach:Use Ohm's law for the series resistor: R = (Vs - Vf) / If. Choosing R equal to or higher than this value keeps current at or below the rating (higher R reduces current and brightness but is safe).
Step-by-Step Solution:
Step 1: Voltage across resistor Vr = 6 - 2 = 4 V.Step 2: Rated current If = 10 mA = 0.01 A.Step 3: Required R = Vr / If = 4 / 0.01 = 400 Ω.Step 4: Any R ≥ 400 Ω keeps current ≤ 10 mA; values below 400 Ω exceed rating.Verification / Alternative check:
Check R = 400 Ω → I = 10 mA (exact). Check R = 200 Ω → I = 20 mA (unsafe). Thus the safe range is 400 Ω and above.Why Other Options Are Wrong:
0–200 Ω: Allows > 20 mA, overstressing LED.200–400 Ω: Still permits up to 20 mA; unsafe at lower end.200 Ω and above: Includes unsafe range below 400 Ω.100–150 Ω: Highly unsafe currents.Common Pitfalls:
Using a resistor value calculated for typical Vf without measuring actual Vf; ignoring LED current derating with temperature.Final Answer:
400 Ω and above