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

R1 = 230 [Ohms]

Pr = 1/8 W

Explanation:

(See attachment)

It´s very useful to draw a circuit to make the problem easier to solve. It´s a simple circuit with a power source of 6V, a resistor R and a LED with nominal values of IF and VF.

IF and VF are the minimum values of current and voltage the LED needs to light on. Therefore our circuit needs to have a current flow of at least 20 [mA] or .02 [A]

We also know the voltage of the power source and the voltage of the diode. Kirchoff´s law says that the sum of all voltages in a closed loop must equal 0. There must be an electrical balance.

Looking at the circuit drawn: The battery has two poles (positive and negative) and current always flows from positive to negative. The LED has anode (+) and cathode (-) and it can only light on if current flows from anode to cathode. We place a + and - in our resistor in that way but we don´t know if it is correct or not. We will know when we apply kirchoff´s law. To apply kirchoff´s law, simply read from the battery source to the right like this:

Notice that if you had placed your resistor poles different, you would´ve gotten a negative voltage in your resistor, that would tell you that the poles are incorrect and you must switch them. This is not the case, we did it right.

Now we know the voltage and current in our resistor. Use Ohm´s Law (V = RI) to find the resistance value you need.

To find the power of your resistor, you can use either one of the next formulas:

You will find that the power dissipated by the resistor is P = .092 [W]

Now we need to select the rating from the ratings available and make sure our power dissipated does not exceed the rating. In this case, 1/8 [W] or .125 [W] is a perfect fit for our application. It exceeds the value of .092 [W] but not by a lot, so you are not oversizing your resistor.

d. learned preferences i believe is your answer

good for her? ? where is the question?