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

Question

A light emitting diode (LED) has a voltage drop of 2V across it and passes a current of 10 mA. When it operates with a 6V 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 mA=10× 10-3A textVoltage=6-2=4V From Ohms law V=IR R=(V/I)=(4/10× 10-3)=400 Ω

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

40 $k Ω$

4 $k Ω$

200 $Ω$

400 $Ω$

The term LED is abbreviated as Light Emitting Diode. It is forward-biased p-n junction which emits spontaneous radiation. Current in the circuit $= 10 m A = 10 \times 10^{- 3} A$ $Voltage = 6 - 2 = 4 V$ From Ohms law $V = I R$ $R = \frac{V}{I} = \frac{4}{10 \times 10 ^{- 3}} = 400 Ω$

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

40 kΩ

4 kΩ

200 Ω

400 Ω

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 Voltage=6−2=4V From Ohms law V=IR R=IV​=10×10−34​=400Ω

40 $k Ω$

4 $k Ω$

200 $Ω$

400 $Ω$

The term LED is abbreviated as Light Emitting Diode. It is forward-biased p-n junction which emits spontaneous radiation. Current in the circuit $= 10 m A = 10 \times 10^{- 3} A$ $Voltage = 6 - 2 = 4 V$ From Ohms law $V = I R$ $R = \frac{V}{I} = \frac{4}{10 \times 10 ^{- 3}} = 400 Ω$