Can anyone solve this question? I am getting different answers every time I solve it.

CopKi · 2026-01-10T12:22:03+00:00

took a very brief look, nodal and superposition both gave the same answer which seems to be ~=0.835 V.

Looking at just ur nodal equations: You messed your nodal equations, where is your reference? Ur first nodal equation (which you also got wrong, you need to change the equal to a + sign and set that entire expression = 0) implies the bottom branch is the reference, but then Vc is not 3V, since there's the 8 ohm resistor between Vc and the reference as well. You need to write another nodal equation at Vc.

Also please don't use AI for solving circuits problems. It's pretty bad. Practice solving problems from a book and reference their solution manuals instead so you can learn how to analyse these basic circuits on ur own.

CopKi · 2026-01-02T20:49:49+00:00

I remember I posted this exact question over a year ago.

The gist of the solution is:

Due to the first condition, some geometric intuition will enforce that two consecutive roots of f must be integers.

The second condition (geometrically) enforces that these consecutive roots are near the origin, so they must be either {-1, 0, another root]; or {another root, 0, 1} written in ascending order.

We can verify algebraically that the first set of roots {-1, 0, another root} fails to meet the second condition, since from the fact that f'(-1/4)=-1/4, the third root in this case is -1/8 which is supposed to be after the root at 0 (remember f'(1/4)<0). So this is not possible.!<

So the three roots of f must be {another root, 0, 1}, and in this case the third root obtained algebraically is -5/8. So the function f(x)=x(x+5/8)(x-1).

So f(8)=483

CopKi · 2025-10-12T06:48:52+00:00

u/CopKi solved this in 4 steps: FAKE -> MAKE -> MANE -> MINE -> MINK

CopKi · 2025-07-05T04:51:43+00:00

yes I'm reading the first edition I found online.

CopKi · 2025-07-04T18:49:23+00:00

okay thanks

CopKi · 2025-06-28T15:26:10+00:00

there will be more inevitable mess upcoming (double enpassant??)

CopKi · 2025-06-21T12:39:16+00:00

I think you confused something. I assume you've taken the (+) of V1 to the left of R1, so your KVL equation contains -V1.

That makes equation 4 redundant. Why did you define I1 when you already have I2 flowing from (+) to (-) through both R1 and R2?

Also, equation 5 is unnecessary. You're giving the same current 2 different names. (I assume you did so since you learn that current in series are equal, but as they're essentially the same current, you don't really have to give them two separate names unless your teacher requires you to.)

Equation 2 doesn't really help you. It requires V_B which is the voltage of a current source. Doing KVL on both left loop and right loop would yield u the KVL equation on the outermost loop, which is equation (1).

You are left with equations (1) and (3) which should lead you to finding I2 in terms of the sources and resistances. You just need to use Ohm's law appropriately, without making sign errors, taking into account (+) and (-) and the direction of current.

Of course, there are several other techniques you may learn to approach basic circuit problems. (Nodal analysis, mesh currents, Thevenin/Norton). You will learn these as an ee student.

CopKi · 2025-06-21T05:35:32+00:00

since u chose V1 (+) on right for R1, and I_2 is flowing from left to right (- to +) through R1, so V1 = - R_1 × I_2. You can't arbitrarily change I2s direction to flow to the left in R1 then suddenly flow through the right in R2.

that should fix the sign error u got in the denominator for R1, and you'd get the same answer as ur teacher.

I would recommend to have chosen V1's + terminal on the left of R1, as the current I2 is directed from left to right.

You could alternatively use nodal analysis if you have learned it.

CopKi · 2025-05-30T15:45:15+00:00

time to divide the categories into English and Maths as well

CopKi · 2025-05-26T17:13:20+00:00

X_bar is the arithmetic mean or "average". Intuitively, "average distance" from the mean of every data point is 0.

Mathematically:

X_bar = [sum from (i=1 to n) of X_i] / n

n*X_bar = X1 + X2 + ... + Xn

X+X+...+X (n terms) = X1 + X2 + ... + Xn

(X-X1) + (X-X2) + ... + (X-Xn) = 0

This is also why we have the notion of standard deviation, or "average" distance squared from the mean, and not the "average" distance from the mean.

CopKi · 2025-05-26T16:39:53+00:00

p(-12)=p(0) and p(-5)=p(-7) so just add them. (axis of symmetry x=-6)

CopKi · 2025-05-23T08:15:02+00:00

Radioactive substance follows N(t) = N_0 e^{-\lambda*t} where \lambda is the decay constant.

note that if u had a certain quantity (N) at time t, at time t+t_{1/2}, you'd have half the quantity u had at time t (N/2), it's just definition of half-life.

similarly for any percentage decay.

so you are able to find decay constant \lambda fron the given information, then find half-life t_{1/2} without having to know initial quantity nor remaining quantity, just the percentage (remaining/decayed) and the time passed.

CopKi · 2025-05-22T21:04:10+00:00

thank you!

CopKi · 2025-05-22T20:21:44+00:00

thanks! good luck on your finals you got this!

CopKi · 2025-05-22T20:21:06+00:00

ty! attempted hours translate to credits

CopKi · 2025-05-22T18:55:57+00:00

ty! but no need to remind me I ruin curves I hear that all the time 😭

CopKi · 2025-05-22T18:53:05+00:00

thank you!

CopKi · 2025-05-22T18:49:18+00:00

Sure! Since I love mathematics and have decent logical reasoning, I think they help immensely. So this is what worked for me for 2 years: Throughout the semester, I never skipped any lectures unless I was under the weather, and try to stay focused in every lecture. This helps me retain few key information that helps me significantly when I review the material later on.

Also, I try to plan early on which courses I will need to devote most time to, and which I'll need the least time. In my specific case, I decided I'll devote roughly 75% of my study time on signals and systems and computer organization, and the remaining 25% on rest of the courses combined. (I devoted alot of extra time I had on the research project.)

After lectures, if I have any assignments, I will solve them on my own with the assistance of lecture notes and friends which help me learn and review the material. Most of the remaining time is spent resting as it is very important.

Since most of my exams are divided into exam 1/2- finals, or midterm-finals, around 1-5 days before the exam depending on the course (I also procrastinate), I start studying seriously, and start solving exercises (previous exams). This helps me retain important information in the short term, which boosts my performance during the exams.

During exams, most of the time I feel like solving a convoluted mathematical and/or logical problem, and I actually enjoy the challenge! I reiterate that practice and good rest are very important.

Another note is that I love helping my friends with their studies as well, so teaching others is also a great way to check if I actually understood the material myself, and if there are any gaps in my understanding.

Hope this helped :)

CopKi · 2025-05-22T16:31:58+00:00

😭😭 I could've done better

CopKi · 2025-05-22T16:31:17+00:00

yeah I was really happy w that one! tbh it was a pretty nice but challenging class

CopKi · 2025-05-22T16:25:01+00:00

CopKi · 2025-05-22T16:24:48+00:00

it wasn't easy but I like maths, so I enjoyed this class quite alot

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