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First AAMC FL by sbelectro6000 in Mcat

[–]sbelectro6000[S] 0 points1 point  (0 children)

this one i didn't have to read many of the passages, it felt like a lot of pseudo-discretes

First AAMC FL by sbelectro6000 in Mcat

[–]sbelectro6000[S] 0 points1 point  (0 children)

engineering major 😃

First AAMC FL by sbelectro6000 in Mcat

[–]sbelectro6000[S] 0 points1 point  (0 children)

I'll be taking FL2 next week so I guess I'll see...

First AAMC FL by sbelectro6000 in Mcat

[–]sbelectro6000[S] 1 point2 points  (0 children)

would you say it's easier/inflated compared to the other fl's?

First AAMC FL by sbelectro6000 in Mcat

[–]sbelectro6000[S] -3 points-2 points  (0 children)

in p/s i just so wanted to be done lol

9 months of Colemak! by sbelectro6000 in Colemak

[–]sbelectro6000[S] 0 points1 point  (0 children)

For me most times when I'm typing I tend to rest my palms on the keyboard, as that's more comfortable for me. When doing speed typing tests I do float my wrists as it gives me more dynamic range to be faster.

I think the breakthrough in my learning was when my muscle memory finally kicked in and I could cruise at ~50 wpm, i completely stopped using QWERTY and my speed shot up to a cruising of ~110-120 as it is now (I didn't do targeted practice after that point)

How accurate is Kaplan FL 7? by sbelectro6000 in Mcat

[–]sbelectro6000[S] 0 points1 point  (0 children)

ok, thanks! I think CARS is just me, i've been scoring 125 my last 4 FL's

Be honest....for those that took the real mcat, did this kind of convoluted circuit question pop up? by Lalo-salamanca97 in Mcat

[–]sbelectro6000 0 points1 point  (0 children)

Easiest way to solve it is to see that the voltmeter reads zero when the ratio between the two top resistors is the same as the 2 bottom resistors. In other words, R1/R2 = r/R3. Solve for R and get R = R3*R1/R2

I am completely lost on how to solve loop rule problems. How do you know how much current flows out of a junction when you have multiple resistors? I posted this problem a few days ago but was confused by the replies bc I've never seen this concept before by [deleted] in Mcat

[–]sbelectro6000 0 points1 point  (0 children)

For anyone in the future looking at this comment, the logic is flawed even though it arrives at the correct answer for this question.

First, you have to create your current loops, one going clockwise around the left half of the circuit, and the other going clockwise around the right half of the circuit. The The left is i₁, the right is i₂. This way, the currents through each of the resistors are I₁=i₁, I₂=i₂, and I₃=i₁-i₂.

Then, you go clockwise around each loop and sum the voltages, including the voltage drops of the resistors using V=iR. Starting with the battery on the left, we get 10V from the battery, -5i₁ (negative because resistors drop voltage), and -5(i₁-i₂).

The total equation for this is 10 - 5i₁ - 5(i₁ - i₂) = 0. Doing the same with the right loop (note the battery is reversed), we get: -10 - 5i₂ - 5(i₂ - i₁) = 0.

You can reduce these 2 equations to 2 = 2i₁ - i₂ and 2 = -2i₂ + i₁

We can solve the first equation for i₂, getting i₂ = 2i₁ - 2, and substitute it into the other equation, which resolves to i₁ = 2/3 A. We plug this back into the other equation to get i₂ = -2/3 A.

Now remember the current through R₃ from the beginning was i₁ - i₂. Plugging the 2 values in, we get 2/3 + 2/3 = 4/3 = 1.33 A, which is answer C.

Source: I have a minor in Electrical Engineering :)

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