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The Art of Electronics: Solutions to Exercises?
masterquiper:
Hi.
Thanks for your the information shared, thoug i would like to ask some:
1. In a real circuit, the resistor will heat up, how can I calculate that power loss? (Is it then just simply using x times more current from power source, without affecting anything else?)
2. I would like to confirm that by testing, which parts should I get to do even some further exercises practically? (as i am not close to such a shop) I will go and buy today a multimeter and some resistor and led's.
3. Is there still no book with solutions out?
Thanks a lot for your help!
Masterquiper
Nusa:
An online source:
https://www.scribd.com/document/291773195/Student-Manual-For-The-Art-Of-Electronics-pdf
rstofer:
--- Quote from: masterquiper on April 04, 2017, 12:27:28 pm ---Hi.
Thanks for your the information shared, thoug i would like to ask some:
1. In a real circuit, the resistor will heat up, how can I calculate that power loss? (Is it then just simply using x times more current from power source, without affecting anything else?)
--- End quote ---
Back to Ohm's Power Law of which there are a couple of variations:
P = I * E - power dissipation is current through resistor times voltage drop across resistor
P = I^2 * R - power dissipation is current through resistor squared times value of resistor
P = E^2 / R - power dissipation is voltage across resistor squared divided by value of resistor - easiest to measure
Note that changing from one form to another just uses the relationships from Ohm's E=IR law.
--- Quote ---2. I would like to confirm that by testing, which parts should I get to do even some further exercises practically? (as i am not close to such a shop) I will go and buy today a multimeter and some resistor and led's.
--- End quote ---
A bunch of resistors of various values and a battery. The actual values don't impact the learning experience, only the numbers change. In fact, you won't be able to buy resistors with absolute textbook values so you can expect your measurements to be off by a little bit but possibly as much as 10%. Don't forget to measure the battery voltage as you measure the resistor drops because the battery will be dropping supply voltage throughout the process.
Try to keep your resistor values at or above 1k Ohm. The idea is to limit current, reduce heat and save on batteries.
That said, you may need lower values when you try to determine the Thevinin model for a real battery. We do this by measuring the open circuit battery voltage and then measuring the battery voltage after a load is supplied. The change in voltage is due to the internal (Thevinin) resistance. To get this measurement may require a significant load - maybe 100 Ohms, maybe less. A 12V battery with a 100 Ohm load will dissipate 1.4W at the resistor (E^2 / R). Use a lower voltage battery, maybe a single AA.
masterquiper:
Thanks a lot for your replies. Now I can go and get the parts and check the exercises (Looks like i am ready :D :D :D)
I think it looked that strange, as it is quite useless to only connect a resistor, nothing else to a batt.
In this simpler case:
Using 5V as Input (from arduino), adding a resistor of 150 Ohms and a single LED of 2V and 20mA. The resistor will use 0.16amps and the LED 0.02? This looks quite inefficient, or am I wrong?
Really appreciate,
Masterquiper
Nusa:
I can't correct your math, since you didn't show it. No idea where you got 0.16 amps.
Current doesn't work like volts or watts. The resistor and the LED will have the same current.
Google "selecting LED resistor" for explanations in depth on how 150 Ohms was picked and why.
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