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
- A40 k
- B4 k
- C200
- 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 .