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erik [133]
3 years ago
15

Discuss the difference between the observed and calculated values. Is this error? If yes, what is the source?

Engineering
1 answer:
Scrat [10]3 years ago
8 0

Answer and Explanation:

In any experiment, the observed values are the actual values obtained in any experiment.

The calculated values are the values that are measured by using the observed values in a formula.

The observed values are primary values whereas the calculated values are the secondary values as calaculations are made using observed values.

Yes, if the observed values are of low accuracy.

The values should be recorded with proper care and attention in order to avoid any error.

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Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a
nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
  3. Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

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It will really help you thank you.

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