RLC 4-16-2020

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

I S I_S is the current flowing out of the source. 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. Let I S m i n I_{Smin} be the smallest source current value between t = 0 t = 0 and t = t = \infty .

Determine the following quantity:

Q = I S m i n I S 0 I S \large{Q =I_{Smin} \, I_{S0} \, I_{S \infty} }

Details and Assumptions:
1) V S = 10 V_S = 10
2) L 1 = L 2 = 2 L_1 = L_2 = 2
3) C 1 = C 2 = 1 C_1 = C_2 = 1
4) R 1 = R 2 = R 3 = 1 R_1 = R_2 = R_3 = 1
5) All three currents are positive


The answer is 636.66.

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1 solution

Steven Chase
Apr 17, 2020

The state variables are the inductor currents and the capacitor voltages. Derive expressions for the time derivatives of the state variables in terms of the state variables and the forcing function.

Basic equations:

V L 1 = L 1 I ˙ L 1 V L 2 = L 2 I ˙ L 2 I C 1 = C 1 V ˙ C 1 I C 2 = C 2 V ˙ C 2 V_{L1} = L_1 \dot{I}_{L1} \\ V_{L2} = L_2 \dot{I}_{L2} \\ I_{C1} = C_1 \dot{V}_{C1} \\ I_{C2} = C_2 \dot{V}_{C2}

Re-writing the left-hand sides in terms of state variables and the source voltage:

V S R 2 I L 1 V C 2 = L 1 I ˙ L 1 V C 2 = L 2 I ˙ L 2 V S V C 1 R 1 = C 1 V ˙ C 1 I L 1 I L 2 V C 2 R 3 = C 2 V ˙ C 2 V_S - R_2 I_{L1} - V_{C2} = L_1 \dot{I}_{L1} \\ V_{C2} = L_2 \dot{I}_{L2} \\ \frac{V_S - V_{C1}}{R_1} = C_1 \dot{V}_{C1} \\ I_{L1} - I_{L2} - \frac{V_{C2}}{R_3} = C_2 \dot{V}_{C2}

And the source current is:

I S = I C 1 + I L 1 = V S V C 1 R 1 + I L 1 I_S = I_{C1} + I_{L1} = \frac{V_S - V_{C1}}{R_1} + I_{L1}

Then we numerically integrate, as usual. By inspection, the currents at t = 0 t = 0 and t = t = \infty are 10 10 . The minimum current is 6.366 \approx 6.366 . Simulation code is attached.

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import math

# Parameters

VS = 10.0

L1 = 2.0
L2 = 2.0

R1 = 1.0
R2 = 1.0
R3 = 1.0

C1 = 1.0
C2 = 1.0

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

# Initialize Simulation

dt = 10.0**(-6.0)
t = 0.0
count = 0

IL1 = 0.0
IL2 = 0.0

VC1 = 0.0
VC2 = 0.0

IL1d = (VS - R2*IL1 - VC2)/L1
IL2d = VC2/L2

VC1d = (VS - VC1)/(R1*C1)
VC2d = (IL1 - IL2 - VC2/R3)/C2

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

# Numerical Integration

ISmin = 9999999.0

while t <= 40.0:

    IL1 = IL1 + IL1d * dt
    IL2 = IL2 + IL2d * dt

    VC1 = VC1 + VC1d * dt
    VC2 = VC2 + VC2d * dt

    IL1d = (VS - R2*IL1 - VC2)/L1
    IL2d = VC2/L2

    VC1d = (VS - VC1)/(R1*C1)
    VC2d = (IL1 - IL2 - VC2/R3)/C2

    IS = (VS-VC1)/R1 + IL1

    if IS < ISmin:
        ISmin = IS

    t = t + dt
    count = count + 1

    if count % 10000 == 0:
        print t,IS

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

print ""
print dt
print ""
print ISmin
print (ISmin*10.0*10.0)

#>>> 

#0.0001

#6.36650497537
#636.650497537
#>>> ================================ RESTART ================================
#>>> 

#1e-05

#6.36656156622
#636.656156622
#>>> ================================ RESTART ================================
#>>> 

#1e-06

#6.36656722509
#636.656722509
#>>> 

@Neeraj Anand Badgujar Here is my approach, for reference

Steven Chase - 1 year, 1 month ago

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@Steven Chase Sir at step 21 it's written dt=10** (-6) . Does it mean that the phython had broked the time in such small time to calculate accurately?.

A Former Brilliant Member - 1 year, 1 month ago

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Yes, usually I like to run the simulation with several different time steps spanning several orders of magnitude, to see that I get the same result with each value of the time step. For example, we can run with 1 0 4 10^{-4} , 1 0 5 10^{-5} , and 1 0 6 10^{-6} and see that the results are similar all three times.

Steven Chase - 1 year, 1 month ago

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@Steven Chase Thanks, Yes Sir I understand it fully. I upvoted it also .I was making a question but not able to solve. The question is that a charged particle ( q = 1 c q=1c ) is moving in the line z = 1 z=1 moving towards the + x +x axis with a speed 1 m s 1 1ms^{-1} . A loop is placed at x 2 + y 2 = 1 x^{2}+y^{2}=1 and we can take loop's resistance as 1 Ω 1\Omega .Find the rate of current induced in the loop. How can we solve this ?? Please help.

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member I found a nice resource for the magnetic field generated by a moving point charge. If you combine this with the integral form of Faraday's law, you should be able to solve for the voltage induced in the loop. That sounds like a fun problem.

https://www.pdx.edu/nanogroup/sites/www.pdx.edu.nanogroup/files/lecture 9 Magnetic Fields produced by a current ph 222 PRINT_9.pdf

Steven Chase - 1 year, 1 month ago

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@Steven Chase Thanks for sharing link. Yes it's a fun problem but it will require hard calculus I think. What specification did this link have in physics?

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member What do you mean by specification? Do you plan to post the problem?

Steven Chase - 1 year, 1 month ago

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@Steven Chase @Steven Chase Sir I have planned but I think I can't solve it. So was just telling you the problem. So that you can make or post the problem.

A Former Brilliant Member - 1 year, 1 month ago

@Steven Chase @Steven Chase Sir according to you which is best udemy, coursera, or edx??

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member I'm not familiar with any of those, actually

Steven Chase - 1 year, 1 month ago

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@Steven Chase @Steven Chase that flux problem which you posted of multi coloured it was very easy. Can you increase the difficulty?

A Former Brilliant Member - 1 year, 1 month ago

@Steven Chase @Steven Chase Sorry if you feel any inconvenience.

A Former Brilliant Member - 1 year, 1 month ago

@Steven Chase Sir are you planning to post that question of charge moving in + x +x direction with a changing flux in the ring. I am planning but can't able to solve. So you can make and post question. It would be very interesting.

A Former Brilliant Member - 1 year, 1 month ago

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How about this? Instead of calculating the induced voltage in the loop, why don't you post a problem that asks to find the magnetic flux in the loop at a particular time, due to a moving point charge? That problem is considerably easier, and it is still interesting

Steven Chase - 1 year, 1 month ago

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@Steven Chase @Steven Chase Ok Sir i am posting.

A Former Brilliant Member - 1 year, 1 month ago

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