Was just wondering, is it possible to find the elastic limit of a material without a stress-strain graph?
Thanks
I'm not sure if it's possible without SOME knowledge of the behaviour (a graph, a table of stress and strain values, a force/extension graph), like there's certainly no VCE level formula for calculating the elastic limit of a material based on other quantities. There may be some formula that involves a lot of other properties of the material (but I doubt it).
If a resistor or even another diode is placed in series in front of a reversed biased diode (the current flows through the resistor before it hits the reverse biased diode), why doesn't current still flow through. For example why isn't any power dissipated? The current has to go through the resistors and such to get to the reverse biased diode?
Good question. Current in series
has to be the same everywhere. Think of a road with a fixed amount of lanes and lots and lots of cars. The rate of the slowest moving cars determines the rate of
all cars because otherwise fast moving cars would catch up and crash into slower cars. This analogy has plenty of holes but the basic idea is that current cant be high in some sections and low in others if those sections are in series.
So, even when current wants to leave the battery, the electrons can't physically go anywhere because there's a blockage all the way along the wires at the diode. (a reverse biased diode is just like a hole in the circuit - no current flows around then either)
Thus, no current flows through the wires, or the resistor, and that's why P=VI=V*0=0.
What happens to the current in the diode when its in reverse biased. Does it send it back the other way? Does it just store it? Or does it get used 100% as heat energy?
Same explanation as first question, there
is no current in this circuit. Nothing happens with it. If a small amount of current were to flow (all real reverse bias diodes let a tiny leakage current through, i think), it would just go round the circuit like normal - the diode acting like a very high-resistance resistor.
For diode labeling conventions. Are we supposed to always change the polarity of the circuit or can we just draw the diode facing the other way? Does it matter at all? Because sometimes I just don't look at which way the circuit is and assume a diode is in forward bias when it is really in reverse.
Conventional current flows out of the bigger terminal of the battery (the positive one) and if that direction around the circuit points the same way that the triangle points, the diode is forward biased.
Electrons actually flow the other way in a real circuit, but the convention still holds.
Always watch out for trick questions 'what is the current in this circuit' when they sneakily put it in reverse mode.
If you're asked to draw a circuit with the diode in the other mode, it's probably easiest to reverse the diode direction rather then the battery direction: it's a lot easier to tell the difference for the diode. An assessor might not see that you changed the battery.
What exactly determines the switch on voltage of a diode. I know the material but what are the differences in material. Is a diode technically still a resistor? Are there other non-ohmic resistor type things as well?
This is to do with the magical chemistry that makes diodes function so I'll let someone else try to answer it.
BUT for the other bit, a diode can be thought of as a resistor but it has no well defined resistance because it is non ohmic (resistance is not the same for different voltages/currents, V-I graph not linear, etc)
You can calculate R=V/I for any given situation but your result is meaningless for the same diode in a different circuit. It's merely the 'effective resistance' of the diode, if you replaced it with a resistor of that same resistance you would get the same behaviour for that circuit.
Other non ohmic devices studied in VCE include temperature dependant resistors, light dependent resistors, and the other types of diodes (LED's and photodiodes)