well, you could measure the resistance through the battery clip...
Nigel, I used this method to measure the resistance in my spagetti 2.0/It pivots robot, to get 15k. It even has a comparator, and some transistors and some other stuff, so i guess you could!
well, you could measure the resistance through the battery clip...
Nigel, I used this method to measure the resistance in my spagetti 2.0/It pivots robot, to get 15k. It even has a comparator, and some transistors and some other stuff, so i guess you could!
well, if it was extremely low, say around 0 it would indicate a short across the battery. It could also mean that if you had a somewhat low resistance, it could indicate a bad component. also I have done this before on my robot, which i reccomend that you read. It's called Spagetti 2.0/It pivots. You might enjoy reading it!
In order for a short to appear across the supply terminals of that circuit, you would need to have s/c resistors as well as transistors. Not to mention the diodes. I've never ever experienced an s/c resistor and infact I don't think it's possible to acheive with modern resistors.
A better question would be; how much current will this circuit draw?
Since only one, lets assume RED, LED will be on at one time:
(9V - 2V) / 470hm: = 14.9ma
Since each transistor base-emitter junction drops 0.7V and there is a 10K resistor on it:
(9V - 0.7V) / 10,000hm: = 0.83ma
So, 14.9 + 0.83 + 0.83 = 16.56ma
The current draw is an approximation because of the dynamic nature of the circuit. (I did not try to account for the charge/discharge of the capacitors or the Vce voltage drop in the circuit, but it is close enough for hobbyist purposes)
In order for a short to appear across the supply terminals of that circuit, you would need to have s/c resistors as well as transistors. Not to mention the diodes. I've never ever experienced an s/c resistor and infact I don't think it's possible to acheive with modern resistors.
No, I don't think it is? - although I have known very old resistors (in valve circuits) go low resistance - basically getting too hot and the resistance dropping, this causes it to get even hotter and the resistance drops still further.
In order for a short to appear across the supply terminals of that circuit, you would need to have s/c resistors as well as transistors. Not to mention the diodes. I've never ever experienced an s/c resistor and infact I don't think it's possible to acheive with modern resistors.
Brian, I was referring to self solder jobs. It is quite easy to get lost and accidently connect the negative and positive buses together, creating a short circuit. That is what i meant. Sorry for the confusion.
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Exactly as Kchriste said. Calculate the current drawn by the circuit and since you know the voltage supplied to it you'll have your equivalent resistance. Its simply a Thevenin equivalence problem.