Oscillating Circuit

In an oscillating circuit , the coil inductance is equal to L and the capacitor have capacitance C 1 C_1 and C 2 C_2 . The capacitor were charged to a voltage V V , and then the switch S w S_w was closed.
Find the peak value ( I m a x ) (I_{max}) of current flowing through the coil in Ampere.

Details and Assumptions
1) L = 2.5 × 1 0 3 H L=2.5×10^{-3} H
2) C 1 = 2 × 1 0 6 F C_{1}=2×10^{-6} F
3) C 2 = 3 × 1 0 6 F C_{2}=3×10^{-6} F
4) V = 180 V V=180V
5) The wire used in the circuit have 0 0 resistance.

The problem is not original.


The answer is 8.

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1 solution

Steven Chase
Jun 4, 2020

Combined capacitance:

C = C 1 + C 2 C = C_1 + C_2

The energy in the circuit transfers back and forth between the capacitance and the inductance:

1 2 L I m a x 2 = 1 2 C V 2 I m a x = C V 2 L 8.05 \frac{1}{2} L I_{max}^2 = \frac{1}{2} C V^2 \\ I_{max} = \sqrt{\frac{C V^2}{L}} \approx 8.05

@Steven Chase please help me in this problem, I will share my opinion if you say yes.

If I use a small angle approximation, I get the following:

D m i n = 2 cos ( 16 π ϵ 0 m v 2 q 2 ) D_{min} = 2 \ell \cos \Big( \sqrt{\frac{16 \pi \epsilon_0 m \ell v^2}{q^2}} \Big)

Steven Chase - 1 year ago

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@Steven Chase Sir answer according to book
Please share your whole approach. I will share the solution of the book, if you say. I think you solution will be correct because I believe in you more than the book.

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Keeping in mind that the argument of the cosine function is small, my answer is equivalent to this. If you do a two-term Taylor expansion of the cosine, and then do a little more algebraic manipulation, you get the same answer as the book. The solution is fairly involved, so I will post it tomorrow. I am going to bed soon.

Steven Chase - 1 year ago

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@Steven Chase @Steven Chase ok sir Thanks. This is the book solution, which is very small and it seems me incorrect also. In your opinion it is correct or not??

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@A Former Brilliant Member Yeah, it looks right. That approach is simpler than mine, but they both yield the same answer (just in different forms)

Steven Chase - 1 year ago

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@Steven Chase @Steven Chase but in the solution how can we write 2 2 m v 2 \frac{2}{2} mv^{2} I do not think that both the ball have equal velocity equal to v. After that the instant when string is pulled with velocity v. And in this system in my opinion in any time both balls can't have velocity equal to v.
Please correct me if I am wrong

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@A Former Brilliant Member That's what I thought originally. But then I read more closely and saw that they are using a reference frame which coincides with the center of the string. And from that point of view, the masses are moving initially.

Also, do you mind if I use the image from that 6th-order LC problem to post a follow-up problem?

Steven Chase - 1 year ago

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@Steven Chase @Steven Chase please share your solution of that question. Is it necessary that when the closest approach comes the velocity in y y direction will be zero. ????

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@A Former Brilliant Member I have posted my solution in the notes section. Referring back to the book solution, on closest approach, the charges are at rest with respect to the middle of the string. And since the middle of the string is moving upward, I suppose that means the charges are as well.

Steven Chase - 1 year ago

@Steven Chase sir i have learned that how to solve a single integral using midpoint theorem .
before solving double integral i want to solve differential equation.
Which method is best for solving differential equation ? I have taken a simple 1st order differential equation from my maths book .
x d y d x + y = x 3 y 4 x\frac{dy}{dx} +y =x^{3}y^{4}
how can i solve this ?
Thanks in advance .

A Former Brilliant Member - 11 months, 3 weeks ago

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You can isolate the derivative term and then numerically integrate

Steven Chase - 11 months, 3 weeks ago

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@Steven Chase sir btw how can i plot α \alpha vs θ \theta in excel α = 3 2 sin 2 θ 1 \large \alpha = \frac {3 }{2\sin^{2} \theta}-1

A Former Brilliant Member - 11 months, 3 weeks ago

@Steven Chase sir can you please solve the above differential equation and share the code with me after that i will solve 1st differential equations myself .
i wii take your code as an example.
thanks in advance

A Former Brilliant Member - 11 months, 3 weeks ago

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@A Former Brilliant Member If you have an analytical solution, you can use my initial conditions and check my result

Steven Chase - 11 months, 3 weeks ago

You have to be careful with this one, because the function can blow up very easily if not initialized in a sensible way.

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import math

dx = 10.0**(-6.0)

x = 1.2
y = 0.3

yd = ((x**3.0)*(y**4.0) - y)/x

count = 0

#print x,y

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

while x <= 5.0:

    y = y + yd*dx

    yd = ((x**3.0)*(y**4.0) - y)/x

    x = x + dx
    count = count + 1

    #if count % 100 == 0:
        #print x,y,yd

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

print ""
print ""

print dx
print x
print y

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

#>>> 

#0.0001
#5.0
#0.0775400886572
#>>> ================================ RESTART ================================
#>>> 

#1e-05
#5.00000999998
#0.0775502950792
#>>> ================================ RESTART ================================
#>>> 

#1e-06
#5.00000000035
#0.077551476454
#>>> 

Steven Chase - 11 months, 3 weeks ago

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@Steven Chase Thanks sir.
BTW I have noticed that sometimes you write 1000% and sometimes you write 100%
What is logic behind that?

A Former Brilliant Member - 11 months, 3 weeks ago

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@A Former Brilliant Member Sometimes I want to print once for every hundred simulation steps, and sometimes I want to print once for every thousand.

Steven Chase - 11 months, 3 weeks ago

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@Steven Chase @Steven Chase sir can you post a RLC problem.
I will try my best to solve using Python.
Thanks in advance

A Former Brilliant Member - 11 months, 3 weeks ago

@Steven Chase sir are you free now??
I want to ask you 1-2 simple doubts of kinematics.
Thanks in advance.

A Former Brilliant Member - 11 months, 1 week ago

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