Question

A light-emitting diode (LED) is a component that produces light. When an LED has current running through it, we sometimes make the rough approximation that the voltage across the LED is constant regardless of how much forward current is flowing through it (within a certain range of current). This approximately constant voltage is given the designation VF. LED data sheets specify the current to be used through the device when lighting it; this current is given the designation IF.Choose a resistor R1 (specify a resistance AND power rating as your answer) to be used in the following circuit to illuminate a red LED with a VS = 6 V battery. This red LED has IF = 20 mA and VF = 1.4 V. Use the smallest power rating possible for the resistor; you have the following resistor power ratings available: 1/16 W, 1/8 W, 1/4 W, 1/2 W, and 1W. The actual power dissipated by the resistor should not exceed the power rating of the resistor you choose. (You do not need to choose a standard resistance value for the resistor.)

Answer 1

Answer:  230 Ω, 1/8 W

Explanation:

In order to keep a constant forward current of 20 mA, causing a voltage loss of 1.4 V within the LED, we need to choose a resistor, where there must be a loss that allow us to use 2nd Kirchoff's Law as follows:

6V = 1.4 V + 20mA * R  ⇒ R= 4.6 V / 20 mA = 230 Ω.

Now, in order to choose the power rating for thee resistor, we can compute first the power dissipated at the resistor, as follows (Joule's Law):

P= IF² * R = (20mA)²* 230 Ω = 0.092 W.

As the dissipation is almost 0.1 W, we need to choose a resistor which power rating (being the smallest possible) be greater than 1/10 W, so we can choose safely 1/8 W.