Back to school: The start of 6.002x

As I mentioned a couple of weeks back, I signed up for a free online self-study course in Circuits and Electronics, run by MIT under the MITx programme. Today the course started and I wish to say one thing should my Electronics teachers ever see this (David Brooker or Paul Abbot) half of what the course covers I have never seen before.

When it comes to pure maths, my skills are pretty week, fortunately Google has been able to help me answer some of the questions, which so far concern looking at discrete abstractions based on the V-I relationship. From here on, this contains some maths that will help anybody else looking to do these kinds of questions. First, if you haven’t seen these two equations, then leave now, this isn’t the post for you!

Voltage = Resistance * Current (V = RI)
Power = Voltage * Current (P = VI)

This is Ohm’s law and the most basic of the power laws. What about if you only know the Power and Resistance though? Well it turns out that there is in fact a complete set of rules for Power, Voltage, Resistance and Current which are shown below:

Power = Voltage * Current (P = VI)
Power = Resistance * Current Squared (P = VI^2)
Power = Voltage Squared / Resistance (P = (V^2)/R)
Voltage = Resistance * Current (V = RI)
Voltage = Power / Current (V = P/I)
Voltage = Sqrt(Power * Resistance) (V = Sqrt(PR))
Current = Sqrt(Power / Resistance) (I = Sqrt(P/I))
Current = Power / Voltage Squared (I = P/(V^2))
Current = Voltage / Resistance (I = V/R)
Resistance = Power / Current Squared (R = P/(I^2))
Resistance = Voltage / Power (R = V/P)
Resistance = Voltage / Current (R = V/I)

Still following? Good! So we are now able to derive any of the four main units given that we know at least two of them. What if we’re not talking about DC power though, what if we want to do the calculations on AC power? The good news is that you can with a little work. The first thing to consider is that in an AC system, the voltage changes in respect to time, therefore is time is stopped, then the voltage will remain fixed. Therefore the first option is to sample the line voltage at a given time and perform the DC calculations based on the voltage. There is however a catch in that voltages are quoted as RMS when we need the actual line voltage where we need to use the peak voltage (Approximately 325 volts for a 230 volt RMS supply). This is where the following equations help and save the need for doing integration – which I can’t remember!

Peak Voltage = Sqrt(2) * RMS Voltage (Vp = Sqrt(2)Vr)

Where you need to find the power for a complete cycle where the wave is a pure sine waveform, the following equations can be used. Where getting the average voltage as a DC value, the RMS value should be used:

RMS Voltage = 0.707 * Peak Voltage (Vr = 0.707Vp)
Average Voltage = 0.637 * Peak Voltage (Va = 0.637Vp)

Right, that’ll do now. More to follow as I watch more lectures and do the practicals.


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