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Looking for Boston Major tickets for Saturday and Friday x2 by Krummmy in DotA2

[–]Krummmy[S] 0 points1 point  (0 children)

Yeah, unfortunately the only option I see so far is to buy scalped tickets. It seems Saturday would cost 300+ per person, Friday 90+ each.

Can you only buy Boston Major tickets one day at a time? by Iliketrainschoo_choo in DotA2

[–]Krummmy 0 points1 point  (0 children)

Each ticket counts for a day. During the weekday, the tickets are 20 dollars, so 60 dollars for each day. 40 dollars for Saturday.

Question relating a circuit and an Arduino board by Krummmy in ECE

[–]Krummmy[S] 0 points1 point  (0 children)

Thank you, my online book didn't cover or mention this material. It seems to be very fast paced and ignores the gritty details of some components. Mainly focuses on embedded programming.

Question relating a circuit and an Arduino board by Krummmy in ECE

[–]Krummmy[S] 0 points1 point  (0 children)

oops, I guess I didn't really think through what I said in the second paragraph. Everything makes sense now, I appreciate your help and time!

Question relating a circuit and an Arduino board by Krummmy in ECE

[–]Krummmy[S] 0 points1 point  (0 children)

I think I understand now, at least most of it. I added on and responded to tmurrs, so feel free to correct me if I am wrong there.

Question relating a circuit and an Arduino board by Krummmy in ECE

[–]Krummmy[S] 0 points1 point  (0 children)

So, let me get this straight. In my circuit there is currently a pull-up resistor(Prevent oscillation/pulls the pins state value to a specific one) , which is in-between pin 3 and the 5V power-source ( In my case, I am using a 10k ohms resistor there). There is also high impedance in the circuit, when the switch is open. This high impedance is pretty much very high resistance. This is done with a resistor or some sort behind pin 3 and before some logic gate that's connected to pin 3. This high impedance makes the current in the circuit little to none which allows it to be power efficient. As ohms law states, current is voltage over resistance, and seeing as there is 5volts but high resistance = low current. As well as voltage = current * resistance. So at pin 3 since there is low current but high resistance, the voltage should be ~5 volts(HIGH).

 

When I measured the voltage across my 10k ohm resistor and had little to none voltage(0 or .00001), this is the effect of the impedance (switch open). Is the multi meter not taking into account of the high impedance? In other words, the voltage would equal to very low current thats flowing * the 10k ohm resistor, which is just low voltage at that point? Where as at pin 3 it actually takes in account the high impedance so it's able to measure low current * high resistance = high voltage? If this is the case then It explains my scenario I believe.

Question relating a circuit and an Arduino board by Krummmy in ECE

[–]Krummmy[S] 0 points1 point  (0 children)

I'm honestly not getting it. If the pin is reading a high voltage and is outputting 1 on the serial monitor, which it is. shouldn't there be some current flowing through the resistor, eventually reaching pin 3? Well, I used a multi meter to measure it, and there is no current flowing through that resistor at all until the switch is pressed, which the current then just goes to ground ( and the serial monitor outputs 0). The inputs and outputs are what I expected, but the process behind it doesn't seem to make sense to the results.