Recent content by s3a

Let's restart. I did a lot of problems for those (Kirchoff's Laws, and Ohm's Laws too, but that's kind of trivial) and my answers matched the answer keys. (For what it's worth, my studying wasn't using Khan Academy, but still.) If I noticed I didn't understand something, I would just say it. I...

The following is (i) and (ii).: I guess I should also add a (iii) that is the voltage when the switch is closed can be determined by getting ##R_{equivalent}## of the digital input resistance and resistance in parallel to it (0 for the pull-up resistor and the pull-down resistor's resistance in...

"Pullups and pulldowns are there to prevent inputs from being undefined and floating while nothing else is connected to said input pin. As described earlier in this thread, that can cause several types of problems." And, (i) prevents that by having a wire with or without a resistor connecting...

Sorry for the delay. To try to get back to the fundamentals, in short, is the concept of a pull-up or pull-down resistor just a matter of (i) having the input pin be connected to VSS or GND so that there is a clear potential difference to prevent a floating voltage, (ii) having a (pull-up or...

Are you talking about the pull-up-resistor case (and not the pull-down-resistor one)? If so, looking at the Wikipedia diagram for it, the table situation you described seems to me to be like that but without the "to digital input" branch-out, which has its own ground and parallelizes things...

Sorry, I'm probably just slow. :P :( So, the resistor has one end attached to the table top, but the other end is not attached to anything? Also, the AA "battery"/cell is attached to the aforementioned resistor end via the table? If so, ignoring magnetic fields from nearby stuff, I'd say the...

Oh. :P I was trying to do what I thought you said and wasn't thinking in a real-life way. So, it's basically Req2 = Req1 in both cases, but when the switch is open, there is no voltage source powering anything and that's what makes the difference, right? So, generally speaking, is the "trick"...

Okay, good to have the GND thing confirmed. As for the other part, I am using the diagram with the pull-down resistor from the image below (obtained from Wikipedia) (so, not the other image that was from the link that berkeman had given (and so I am not making any reference to R1 nor R2 from...

Switch closed: V = R*I V = 0*I V = 0 (logic low) Switch open: V = R*I V = (R1+R2)*I (logic high since R1+R2 is a finite resistance so the voltage source can provide however much current is needed?) So, is the key to recognize the wire of the switch and the resistance of the MCU (R2) as being...

So, I find Req at the pin? Switch open: Req = 1 / (1/Rs + 1/R2) Req = 1 / (1/INF + 1/R2) Req = 1 / (0 + 1/R2) Req = R2 Switch closed: Req = 1 / (1/Rs + 1/R2) Req = 1 / (1/0 + 1/R2) Req = 1 / (INF + 1/R2) Req = 1 / (INF) Req = 0

Actually, I realized that that explanation I gave doesn't address the voltages, so that logic wouldn't be good enough for at least CMOS technology. I know about Ohm's and Kirchoff's laws. In the following image, when BUTTON is pressed, are R1, VCC and input pin in series with the GNDs...

Thanks for your responses and sorry for my late one. It will probably seem trivial to you guys, but this (Reddit) post explains and solves my biggest confusion.: Namely, "What I don't get, is when the button is pressed down. Now, the voltage from the pull-up resistor can go either to ground...

To Baluncore: My confusion was far more basic than that, but thanks anyway. For one, I didn't know TTL was the old stuff and CMOS the new. I did see the word CMOS in a textbook that I will eventually read when my knowledge level rises sufficiently. To both (Baluncore and berkeman): Having said...

For the attached image (from https://en.wikipedia.org/wiki/Pull-up_resistor), could someone please help me intuit what happens in each arrangement when the switch is on and when it's off? So, apparently the pull-up or pull-down resistor is chosen optimally so as to be too low to significantly...

Okay, so, in short, if there are at least two sources or at least one op amp, then polarity matters, right? What about if there is one voltage source and one current source? Polarity still matters then, right? Also, the circuit of this thread only has one source, right? I ask because that's...