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

Question

A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R. The value of R is : (A) 40 kΩ (B) 4 kΩ (C) 200 Ω (D) 400 Ω

Tardigrade Admin · Accepted Answer

The term 'LED' is abbreviated as 'Light Emitting Diode'.
It is forward-biased

p - n

junction which emits spontaneous radiation.
Current in the circuit

= 10 m A = 10 \times 1 0^{- 3} A

and voltage in the circuit

= 6 - 2 = 4 V

From Ohm's law.

V = I R

∴ R = \frac{V}{I} = \frac{4}{10 \times 1 0 ^{- 3}} = 400 Ω

Q. A light emitting diode $(L E D)$ has a voltage drop of $2 V$ across it and passes a current of $10 m A$ . When it operates with a $6 V$ battery through a limiting resistor $R$ . The value of $R$ is :

$40 k Ω$

$4 k Ω$

$200 Ω$

$400 Ω$

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

40kΩ

4kΩ

200Ω

400Ω

The term 'LED' is abbreviated as 'Light Emitting Diode'. It is forward-biased p−n junction which emits spontaneous radiation. Current in the circuit =10mA=10×10−3A and voltage in the circuit =6−2=4V From Ohm's law. V=IR ∴R=IV​=10×10−34​=400Ω

Q. A light emitting diode $(L E D)$ has a voltage drop of $2 V$ across it and passes a current of $10 m A$ . When it operates with a $6 V$ battery through a limiting resistor $R$ . The value of $R$ is :

$40 k Ω$

$4 k Ω$

$200 Ω$

$400 Ω$