Internal resistance and delivering the maximum power to a load resistor

I was teaching this on Friday but ran out of time to go through the calculus with my class, so here is a quick(ish) video about why you get the maximum power delivered when you match the load resistance to the internal resistance. Hopefully you find it useful! https://youtu.be/nEY3OsCPZtU If there’s anything you need helpContinueContinue reading “Internal resistance and delivering the maximum power to a load resistor”

A Level revision timetable!

I’ve written a spreadsheet that creates a revision timetable automatically – just plug in the topics and off you go. It includes interleaving and repetition to promote long term retrieval. Here’s a screenshot (the colours are for broad topics): This is available to enrolled A Level tutees, and there’s a GCSE version available to enrolledContinueContinue reading “A Level revision timetable!”

Looking inside the HP 9820A calculator

I was very kindly and generously gifted a Hewlett-Packard HP 9820A calculator by a good friend and ex-colleague. Unfortunately, the thing does not work, but I will try to get it working when I find the time between other things I have to do. The HP 9820A was the world’s first calculator that used anContinueContinue reading “Looking inside the HP 9820A calculator”

More fun with op amps: differentiation

Yes, I finally managed to get an op amp to differentiate an input potential that varies over time. The top trace is the input potential. The bottom trace is the output from the op amp, showing the first derivative of the input. Huzzah! I connected the differentiator circuit to a sinusoidal input potential, and afterContinueContinue reading “More fun with op amps: differentiation”

Fun with Op Amps

One can see the same potential applied to both inputs of the op amp, causing 0V output potential. That made sense. When connecting the inverting input pin to half the potential of the output (using two 10 kΩ fixed resistors as a potential divider), the output potential became 10V. Well, half of 10V is 5V, so that would make the inverting input potential 5V, the same as in the first case, yet the output potential is different.