The value of current limiting resistor for a stack of 4 LED's connected in series will be ______ if the LED's are 3 V, 3 mA and DC source is 15 V.

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LPSC (ISRO) Technician B (Electronic Mechanic): Previous Year Paper (Held on 4 March 2018)
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  1. 10 Ω
  2. 100 Ω
  3. 1 kΩ
  4. 10 kΩ

Answer (Detailed Solution Below)

Option 3 : 1 kΩ
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Detailed Solution

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Given: 

Source voltage = 15 V

Voltage of LED = 3 V

Current of LED = 3 mA

Now,

The given circuit can be drawn as

F1 Neha 18.1.21 Pallavi D1

The current through the LED circuit must be 3 mA and the voltage drop across them is 3 V each.

Therefore, by applying KVL, we get,

15 – R (3 mA) - 3 - 3 - 3 - 3 = 0

\(R = \frac{{15 - 12}}{3}~k{\rm{\Omega }} = 1~k{\rm{\Omega }}\)

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