Just work through it with whatever material you have. Whether TPR, BR, EK, Kaplan or material online, they all give you what you need to know for the MCAT.
Try to prepare a lesson plan to teach to a class, and run through how you are going to explain each concept. It's a bit of role-playing, with the idea being that it's meant to force you into taking good, condensed notes.
This works for any other concept you are trying to master. The process of absorbing, processing, and synthesizing helps to cement it in your head.
If that sounds a bit awkward for you, then just solve EM problems, open-book. This forces you to identify what you do and don't know, as well as understand the concept or idea in order to solve the problem.
How about an example?
Random EM problem I found online:
5. Two resistors, 2.0 ohm and 3.0 ohm, are connected in parallel with a 12 V battery. Calculate: (a) the effective resistance of the circuit; (b) the current leaving the battery; (c) the current through each resistor; (d) the voltage drop across each resistor.
Okay, so we are dealing with circuits, resistors, and a parallel system. We are also given voltage and ohms. We know a basic equation that gives the relationship between some of these variables, which is V=IR. Great. I don't know what parallel circuits are, so let's look it up. Looking up parallel circuits online tells me they are circuits where current branches, and the total current is the sum of each individual component current (Itot=I1+I2...). Also, it tells me the voltage to each branch is same as the source (Vtot=V1=V2...), and the relationship of resistance is 1/Reff=1/R1+1/R2... Okay great, so far we've acquired quite a bit; let's see if we can begin solving the problem.
First, list what we know: Two resistors, 2.0 ohm and 3.0 ohm, are connected in parallel with a 12 V battery.
We can sum as: R1=2 ohm, R2= 3 ohm, V=12 V.
Now the questions:
(a) the effective resistance of the circuit
From an earlier equation, we can calculate that easily 1/Reff=1/2+1/3 ohm -> 1/Reff=5/6; Reff=6/5 ohm
(b) the current leaving the battery
We have V=IR->I=V/R; we are given V, and we just solved for R. We can think R must be Reff, since we are calculating for total current. I=12V/(6/5)ohm=10 Amps
(c) the current through each resistor;
Okay. This is getting a bit tough. Let's think back on how current behaves in our parallel system. Total current should equal the sum of each individual current, or Itot=I1+I2... Then this means we can calculate each current separately. So I1=V/R1, and I2=V/R2. We remember that V is the same across all branches, so we keep it constant.
I1=12v/2ohm=6 A
I2=12v/3ohm=4 A
(d) the voltage drop across each resistor.
A freebie! We (now) know, that voltage across each branch is same as voltage from source. So 12 v for each branch.
I got the information and problem from a "homework help" website. To see if I can find quality information on a more generalized website, I searched parallel circuits on wikipedia, and I am seeing that the information is presented just as clearly, with better diagrams and neat tricks/mnemonics.
Pretty cool. Now I know the basics of solving parallel circuits. No fair, you say, you already knew how to solve them! Actually, it's been about 4 years since I last touched an EM problem. What if the problem was more complicated, in a XYZ fashion? Then I'd just simply look up what I'd need to adjust with XYZ, reading about everything I don't understand. Okay, this is fine, but I'll forget everything! No sweat, just do more of the same type of problems, this time closed-book. You'll be forced to recall everything and stamp it into your long-term memory.
I suggest though, unlike what I did here, start with the basics and slowly build up. Having a solid foundation in the fundamentals will give you confidence to tackle harder problems.
Do this with everything you don't understand, and soon you'll too know Kung fu.
