Question

what value of series resistance required to limit the current through led to 0.02A with a forward voltage drop of 1.6V when connected to 10v supply

a)420 ohm
b)300 ohm
c)42.2ohm
d)520 ohm​

Answer 1

Answer:

Explanation:Light emitting diode is a forward biased P-N junction which emits light.

The voltage across it = 2 v

Battery voltage = 6 v

Hence total voltage in the circuit = V = 6 - 2 = 4 v

since a LED has 0 resistance in the forward biased region, total resistance in the circuit

= 0 + r = r

current I = 10 mA = 10/1000 A = 0.01 A

we know that,

R = V/I

=> r = 3/0.01 = 300Ω

Hence the value of the limiting resistor is 400Ω