IV characteristic for a diode

The mark schemes vary on the points they cover. Here's the most recent picture:

If you put a large negative voltage on a diode, eventually you get breakdown. It conducts almost perfectly ie it's as if nothing is getting in the way of the electricity so the current is very high.

IV characteristic for filament bulb

The most recent mark scheme makes these points:

• For low voltages, the line is straight so current is proportional to voltage
• As the current increases, the filament wire heats up
• This increases the resistance
• Increasing the voltage will push more current through, but because the resistance has increased, the current will not go up as much for an equal increase in voltage
• The same thing happens if you put current through the bulb in the opposite direction so the characteristic is symmetrical.
  

Link to Upper Sixth site

This is for anyone from the Upper Sixth using the Lower Sixth blog to revise for the Synoptic Paper:

www.brockostressline.blogspot.com

Baryon number and lepton number

Baryons are not fundamental particles because they are made of smaller pieces (quarks).

A baryon is made from 3 quarks. Baryon number is conserved in an interaction. In other words, if you have a baryon before the interaction, then you need to have a baryon after the interaction.

The odd bit is that an antiparticle like an antiproton has a baryon number of -1. So if a proton and an antiproton interact, the overall baryon number is +1 + (-1) = 0, so the total baryon number after the interaction must be zero as well.

Leptons ARE fundamental particles. They are not made of smaller particles. Lepton number is conserved in an interaction.

Strangeness is conserved in the strong and electromagnetic interactions. Strangeness is strictly not conserved in the weak interaction but the exam keeps coming up with the SPECIAL CASE where you have zero strangeness before and zero strangeness after, so the mark scheme says strangeness is conserved, even though it's a weak interaction.

EMF

As far as the exam is concerned, you are best to define the EMF as the battery output voltage when there is ZERO current through the battery.





Once the current starts to flow round the circuit, current also has to pass through the INTERNAL RESISTANCE of the battery, heating it up. This wastes energy so the final output p.d. of the battery is reduced.

This equation sums up the paragraph above. It says that the output voltage is equal to the EMF minus the p.d. lost heating up the internal resistance.

This versionof the equation is NOT on the data sheet.