Circuits Problem Solution

This problem was sent to me recently. I have made some additions to the graphic.

On every time step of a numerical solution, I do the following:

1) Update the capacitor voltage based on the capacitor current from the previous time step
2) Solve for the voltage at point B based on the nodal equations at point C.
3) Solve for the capacitor current

Assume that point C is the reference voltage point, with zero voltage. Let VC V_C denote the voltage across the capacitor. The node equation at point C is:

0VBR+0(VB+VCV)2R+0+2V(VB+VC)R=0\frac{0 - V_B}{R} + \frac{0 - (V_B + V_C - V)}{2 R} + \frac{0 + 2V - (V_B + V_C)}{R} = 0

Note that VA=VB+VCV V_A = V_B + V_C - V . Solve the above equation for VB V_B . Then find the current through the capacitor as follows:

VA=VB+VCVIC=VBVAR+VB0R V_A = V_B + V_C - V \\ I_C = \frac{V_B - V_A}{R} + \frac{V_B - 0}{R}

Simulation code below. Results are printed at the end of the code.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
import math

# Constants

R = 1.0
C = 10.0**(-6.0)
V = 8.0

us = 10.0**(-6.0)

dt = 10.0**(-10.0)

################################

# Initialize simulation

VC = 0.0

right = (VC - V)/(2.0*R) - 2.0*V/R + VC/R
left = -2.0/R - 1.0/(2.0*R)

VB = right/left

VA = VB + VC - V

IC = (VB-VA)/R + VB/R

VCd = IC/C

################################

# Run simulation

t = 0.0

while t <= (5.0/8.0)*us:

    VC = VC + VCd*dt

    right = (VC - V)/(2.0*R) - 2.0*V/R + VC/R
    left = -2.0/R - 1.0/(2.0*R)

    VB = right/left

    VA = VB + VC - V

    IC = (VB-VA)/R + VB/R

    VCd = IC/C

    t = t + dt

################################

print dt
print t
print ""
print IC
print VA


#>>> 
#1e-08
#6.3e-07

#5.79176835692
#2.55205791077
#>>> ================================ RESTART ================================
#>>> 
#1e-09
#6.25e-07

#5.88135906014
#2.52966023496
#>>> ================================ RESTART ================================
#>>> 
#1e-10
#6.25e-07

#5.88560014166
#2.52859996458
#>>> 

#ElectricityAndMagnetism

Note by Steven Chase
11 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

@Steven Chase Thanks for the solution.

@Steven Chase can we solve this using pen and page only??

Log in to reply

I would like to see the official solution. The numerical solution seems much more natural to me, as usual. I also think it is interesting that they are asking for approximations.

Steven Chase - 11 months ago

Log in to reply

@Steven Chase the official solution is not available in website , in that image you can see it is written NA which means not available.
I will say my physics teacher to upload the solutions now.

Log in to reply

@A Former Brilliant Member Is this your schoolwork or a problem from a university textbook?

Krishna Karthik - 11 months ago

@A Former Brilliant Member If this is your schoolwork, I must say your school is quite hard.

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik @Krishna Karthik this is work of a problem like of JEE.
But it is not in our school course .
I have separately joined a Institute (coaching) for JEE preparation.

Log in to reply

@A Former Brilliant Member Ah. that explains it. Holy shit... JEE's hard. Lol I live in Australia so I can't try JEE. I'm doing IB physics Higher Level; it's sort of in-between JEE and AP Physics.

Bro, you'll definitely get into IIT though. You're already way ahead lol

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik @Krishna Karthik i can give jee advance papers. Do you want to try those problems.??
I have jee level millions of problems.

Log in to reply

@A Former Brilliant Member Sure! Thank you so much bro. But only classical mechanics. I haven't started on Electromagnetism yet lol

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik @Krishna Karthik no problem. I have separate practice problem for each topic. . But the condition is you have to use ony Pen and page.

Log in to reply

@A Former Brilliant Member Yes; of course. I'll try my best. Beautiful!

Krishna Karthik - 11 months ago

@A Former Brilliant Member It'll be amazing practice for my physics exams.

Krishna Karthik - 11 months ago

@A Former Brilliant Member Did you try my new problem btw? I have to see if my solution's right...

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik @Krishna Karthik did you have studied Thermodynamics??

Log in to reply

@A Former Brilliant Member No, not yet. I am working mostly on my maths and computer science; physics is kind of my side thing that I teach myself on brilliant.

This is because I haven't formally started physics in school yet.

I have started on Special Relativity though lol. It's easier than thermodynamics for sure, but I might start on it soon.

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik @Krishna Karthik so basically you want mechanics. I am giving you 27 problems.
2-3 are not in mechanics you can neglect them .

Log in to reply

@A Former Brilliant Member Thanks! Epic bro.

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik @Krishna Karthik Brilliant is not accepting PDF. Let me you 4-5 photos.

Log in to reply

@A Former Brilliant Member You can upload the photos onto a discussion. Photos are fine. Thx mate!

Krishna Karthik - 11 months ago

@Steven Chase sir the hand made solution is ready.
In the first page (in our school, we have been taught that you don't have to derive left in every question)
But for explaining I have derived for you.
My friends always follow school teachers and use this method and I always solve differential equation for this problems .
What do you think which method is best ?
BTW the official solution are not yet uploaded in the website, I solved 1st problem myself and here is the solution below.

Log in to reply

Generally speaking, differential equations and ground-up mathematics is the correct way to do physics. But for high-school or on a test it'll take time to derive, so I just use formulas for high-school.

Krishna Karthik - 11 months ago

And literally all laws in physics are written in the form of ODEs and PDEs, so high-school just uses approximations or makes some values constant so that rates of change and calculus doesn't have to be used. For obvious reasons as not everyone is a maths wiz.

Krishna Karthik - 11 months ago

@Steven Chase sir did you have seen this type of method.

Log in to reply

That's kind of a nice approach. It just depends on what tools you're allowed to use

Steven Chase - 11 months ago

By the way, I'm planning to make a time-domain version of your maximum power transfer problem, if that's OK with you

Steven Chase - 11 months ago

Log in to reply

@Steven Chase did you have woke up now??

@Steven Chase why not sir? It will be interesting.

Log in to reply

Thx! Cheers

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik you can print it out. It will be helpful.

@Steven Chase is it uploaded now??

Log in to reply

Yes, it is

Steven Chase - 11 months ago

Log in to reply

Hmmm... I wonder... Why has Neeraj Anand Badhgujar deleted his account?

Krishna Karthik - 11 months ago

Log in to reply

@Krishna Karthik There was a bit of drama earlier that I won't describe in detail. I don't think he needed to delete his account though.

Steven Chase - 11 months ago

Log in to reply

@Steven Chase What happened though? He was a friend. I hope you won't mind telling me... it's alright if you won't though... thanks

Krishna Karthik - 11 months ago

@Steven Chase I am getting these 2 equations
x and y are charges And Wolfram has failed to solve and I don't think I can solve this with laplace.
So I am planning to use python
But I didn't know how to use python for this work. Can you please little bit guide me?

Log in to reply

On every time step, you can do:

1
2
3
4
5
6
x = x + xd*dt 
y = y + yd*dt 
xd = xd + xdd*dt 
yd = yd + ydd*dt 
xdd = (insert expression here) 
ydd = (insert expression here) 

Steven Chase - 11 months ago

That is just a general format. You can modify as necessary

Steven Chase - 11 months ago

Log in to reply

Thanks, but how to solve this differential equations??

×

Problem Loading...

Note Loading...

Set Loading...