Note: I only need help with E and F. Thank you! Electronic devices that run on b

Question

Note: I only need help with E and F. Thank you!

Electronic devices that run on batteries have a fixed value for the voltage the battery supplies. For example, many modern batteries have DVb, = 3.7 V. Some circuit elements in the device may require a smaller voltage to run. The circuit shown to the right, sometimes described as a “voltage divider” is used to take a fixed voltage source, and provide a fraction of that voltage between the points A and B.

Note: In this problem, assume the battery is ideal. However, if the battery is not ideal, the internal resistance will just add to R1.

B) Of course to use this circuit, you’re going to connect other stuff, modeled here as a “load” resistor RL, between the points A and B. When you do this, will the voltage over the load be larger, smaller, or the same as DVOut = 2.5 V? Explain.

ANSWER: When there's a load resistance:

If we compare this expression with the previous one we can see that we are adding a Rl factor on the denominator so the voltage will be smaller than Vb.

C) You want to design the circuit such that there is a small change in DV2 = DVLwhen RL is added. For this to be true, should RL be much smaller than R2, much larger than R2, or about the same size as R2? Explain.

What makes designing this circuit difficult, is that the value of RL will change depending on what the device is doing. So, if the load is “off”, RL will be infinite, like there’s not anything connected between points A and B. If the circuit is doing something, RL will have some finite value, that may change depending on what is happening.

ANSWER: If you want Vb=Vl The situation must simulate Rl tendind to infitite just as if it doesn's exists so Rl must be greater than R2

D) If R1 = 1 kW (1000 Ohms), what should R2 be so that DVL = DVOut = 2.5 V when RL is infinite (no load resistance)?

ANSWER: If there's no load resistance we can use

E) Using your result from (D), what is the smallest value of RL such that DVL is within 10% of DVOut = 2.5 V. In other words, what is the smallest value of RL such that
DVL > 2.25 V?

F) A problem with this circuit is that power is wasted in the resistances R1 and R2, when you want the power supplied to RL. Using the values you found in (D) and (E), determine the electrical power converted to other forms of energy in R1, R2, and RL.

Explanation / Answer

(E)

The change in Vout allowed = 0.25 V

this will be additional voltage drop across R1 due to increase of current

The increase in current = 0.25/1000 A

This additional current shall flow through Rl so that the voltage across the load = 2.25V

we have (0.25/1000)xRl = 2.25

Rl = 9000 ohms

F)

Power decipated in R1 = (3.7-2.25)^2/1000 = 2.1 mw

Power decipated in R2 = 2.25^/2083 = 2.4 mw

Power decipated in Rl = 2.25^2/9000   = 0.56 mw

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