RLC Circuit (2-11-2020)

A DC voltage source excites an RLC network as shown below. At time t = 0 t = 0 , the inductor and capacitors are de-energized.

Let I S I_S be the current flowing out of the source. Let I S m a x I_{Smax} and I S m i n I_{Smin} be the largest and smallest source current values over all time. Let I S 0 I_{S0} be the source current at time t = 0 t = 0 , and let I S I_{S \infty} be the limiting value of the source current as the elapsed time approaches infinity.

Determine the following ratio:

I S m a x + I S m i n I S 0 + I S \frac{I_{Smax} + I_{Smin}}{I_{S0} + I_{S \infty} }

Details and Assumptions:
1) V S = 10 V_S = 10
2) L = 1 L = 1
3) C 1 = 1 C_1 = 1
4) C 2 = 2 C_2 = 2
5) R 1 = 1 R_1 = 1
6) R 2 = 2 R_2 = 2
7) Three of the current values are positive, and one is negative


The answer is 1.362.

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1 solution

Karan Chatrath
Feb 11, 2020

Let the currents through the branches be as such:

  • Source current: I S I_S
  • Through R 1 R_1 : I R 1 I_{R_1}
  • Through R 2 R_2 : I R 2 I_{R_2}
  • Through C 1 C_1 : I C 1 I_{C_1}
  • Through C 2 C_2 : I C 2 I_{C_2}
  • Through L L : I L I_{L}

  • The charge on C 1 C_1 : Q 1 Q_1

  • The charge on C 2 C_2 : Q 2 Q_2

Therefore, according to Kirchoff's current law:

I S = I C 1 + I R 1 I_S = I_{C_1} +I_{R_1} I S = I L + I R 2 I_S = I_{L} +I_{R_2} I L = I C 1 + I C 2 I_L = I_{C_1} + I_{C_2}

When viewing the circuit as it is, I C 2 I_{C_2} is assumed to be moving from right to left. From the junction of the resistors to that between C 1 C_1 and the inductor.

According to Kirchoff's voltage law:

V S + Q 1 C 1 + L I ˙ L = 0 -V_S + \frac{Q_1}{C_1} + L \dot{I}_L=0 I R 1 R 1 + Q 2 C 2 Q 1 C 1 = 0 I_{R_1}R_1 + \frac{Q_2}{C_2} - \frac{Q_1}{C_1}=0 Q 2 C 2 + L I ˙ L = I R 2 R 2 \frac{Q_2}{C_2} + L \dot{I}_L = I_{R_2}R_2

Initial conditions:

I L ( 0 ) = Q 1 ( 0 ) = Q 2 ( 0 ) = 0 I_L(0) = Q_1(0) = Q_2(0) = 0

This system of equations can obviously be re-arranged into a state-space form but I simply numerically integrated them using a script of code. The variation of source current with time is as follows:

The answer is 1.3618 \boxed{1.3618} . The source currents initially and after a long time can be simply verified by inspecting the circuit. The initial source current is 5 A 5A and the final source current is 3.3333 A 3.3333A .

@Karan Chatrath Sir how you solve these three linear differential equation as I am class 11th student i can't solve this. I typed it on wolfram alpha from there also I can't get any result. So where do i can solve the differential euation. Is there any online technology to solve this which you use.??? Please

A Former Brilliant Member - 1 year, 3 months ago

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I use a computer to solve them using technique called numerical integration. These can also be solved analytically by rearranging the equations but that involves a significant amount of work.

Karan Chatrath - 1 year, 3 months ago

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Can i solve this equation in my mobile also?? I have Redmi 7 pro

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member I don't know. I am not familiar with any tool equipped within a smartphone that can handle numerics. Google might help you find an answer to this.

Karan Chatrath - 1 year, 3 months ago

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@Karan Chatrath In computer from where I will get this numerical integration technique???

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member Do you have basic programming knowledge or are familiar with Excel? You can do these calculations on either platform

Karan Chatrath - 1 year, 3 months ago

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@Karan Chatrath I have no knowledge of programming but familiar with excel

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member That is nice. I suggest you look up some basics of numerical integration. There is plenty of online material. You can easily implement the techniques you learn in Excel. You might even find a tutorial on how to perform numerical integration using Excel, on youtube.

The most basic form of numerical integration is the Explicit Euler's method.

Karan Chatrath - 1 year, 3 months ago

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@Karan Chatrath Sir can I learn to solve this integration myself. Because in solving question I want to solve whole question with myself only if I use any resources I will have false feeling that I have solved the question. I want pure feel.

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member Before attempting to solve such a circuit in a way that you prefer, I suggest that you start with a simpler problem. Consider a series RLC circuit connected to a DC-voltage source. The capacitor and inductor have zero energy at time t = 0 t=0 . Find the current flow through the circuit as a function of time and also find the heat dissipated by the resistor.

Karan Chatrath - 1 year, 3 months ago

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@Karan Chatrath Sir I have done all that . The problem with me is that in every Steven chase sir RLC question I stuck only in maths. Every time I have correct equations.

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member To solve these circuit equations rigorously, the maths is actually simple but extremely tedious. That is why I avoid it despite knowing it. The equations you obtain using KVL and KCL basically need a lot of re-arrangement. I have posted exact solutions for other questions like these. If you are obtaining the equations correctly, I would say that is adequate if your objective is to prepare for an exam such as JEE. JEE will never test your math brute-force skills in the physics section. It's all about conceptual clarity.

If you attempt to solve these questions out of pure love for the subject, and still want an exact solution, I'm afraid there is no shortcut. There are some nice tricks but they all involve a lot of work.

Karan Chatrath - 1 year, 3 months ago

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@Karan Chatrath Yes you are right. Sir can you please suggest me gook book to solve physics beautiful and tricky problems. ??

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member

  • I think you must be familiar with HC Verma's book. The conceptual content and problems are nice there. This is the most recommended source for interesting exercises, and quite rightly so.

  • Apart from that, I would recommend the book by IE Irodov. Its a book containing only unsolved problems which are of a higher level of difficulty from a conceptual viewpoint. And all the answers to problems are exact (closed-form solutions) which you prefer.

  • I would also recommend ' Physics for Scientists and Engineers by Tipler and Mosca'. Google search for the exact name. I hope your school library has a copy, cause it is expensive. There is a large variety of problems there with varying levels of difficulty.

  • Karan Chatrath - 1 year, 3 months ago

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