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NEET2006PhysicsSemiconductorsActual

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

Options

  1. A40 k
  2. B4 k
  3. C200
  4. D400 .

Correct answer

D. 400 .

Step-by-step solution

As LED is connected to a battery through a resistance in series. The current flowing, 10 ~mA is the same. The voltage drop across LED =2 ~V As the battery has 6 ~V , the potential difference across R=4 ~V . i R=4 ~V R= 4 ~V 10 10⁻³ ~A =400 .

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