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 6 V battery through a limiting resistor R , what is the value of R (in ohm)?

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 6 V battery through a limiting resistor R , what is the value of R (in ohm)?,

MARKS App · Accepted Answer

Current in the circuit = 10 mA = 10 × 10 - 3 A and voltage in the circuit = 6 - 2 = 4 V from Ohm's law, V = I R ∴ R = V I = 4 10 × 10 - 3 = 400 Ω

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