RLC 11-23-2020

Electricity and Magnetism Level 4

An exponentially decaying voltage source supplies an R L C RLC circuit as shown. At time t = 0 t = 0 , the inductor and capacitor are de-energized. What is the product of the biggest positive and biggest negative current values which occur in the circuit?

Details and Assumptions:
1) V S ( t ) = e − t V_S(t) = e^{-t}
2) R = L = C = 1 R = L = C = 1
3) The expected answer is negative


The answer is -0.0605.

Steven Chase by Steven Chase , 37, USA

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1 solution

Hosam Hajjir
Nov 23, 2020

Writing the loop equation of in the s s -domain,

V S ( s ) = 1 1 + s = I ( s ) ( 1 + s + 1 / s ) V_S(s) = \dfrac{1}{1+s} = I(s) (1 + s + 1/s)

Therefore, I ( s ) = s ( 1 + s ) ( s 2 + s + 1 ) I(s) = \dfrac{s}{(1 + s)(s^2 + s + 1) }

Using partial fraction expansion, the Laplace transform of the current can expressed as,

I ( s ) = s ( 1 + s ) ( s 2 + s + 1 ) = A s + 1 + B s + C s 2 + s + 1 I(s) = \dfrac{s}{(1 + s)(s^2 + s + 1) } = \dfrac{A}{s + 1 } + \dfrac{B s + C}{ s^2 + s + 1 }

Multiplying both sides of the last equation by ( s + 1 ) ( s 2 + s + 1 ) (s + 1)(s^2 + s + 1) ,

s = A ( s 2 + s + 1 ) + ( s + 1 ) ( B s + C ) s = A (s^2 + s + 1) + (s + 1)(Bs + C)

Equating coefficients of s 2 , s , 1 s^2 , s, 1 in the above equation, we get,

0 = A + B 0 = A + B

1 = A + B + C 1 = A + B + C

0 = A + C 0 = A + C

so that, C = 1 , A = − 1 , B = 1 C = 1, A = -1 , B = 1 , and hence,

I ( s ) = − 1 ( s + 1 ) + ( s + 1 ) ( s 2 + s + 1 ) = − 1 ( s + 1 ) + ( s + 1 2 ) + 1 2 ( ( s + 1 2 ) 2 + 3 4 ) I(s) = -\dfrac{1}{(s + 1)}+ \dfrac{(s + 1)}{(s^2 + s + 1) } = -\dfrac{1}{(s + 1)} + \dfrac{(s + \frac{1}{2})+\frac{1}{2}}{((s + \frac{1}{2})^2 +\frac{3}{4}) }

Using the well-known inverse Laplace transforms on the right hand side, we get,

I ( t ) = − e − t + e − t / 2 ( cos ⁡ ( 3 2 t ) + 1 3 sin ⁡ ( 3 2 t ) ) I(t) = - e^{-t} + e^{-t/2} ( \cos(\frac{\sqrt{3}}{2} t ) + \frac{1}{\sqrt{3}} \sin(\frac{\sqrt{3}}{2} t ) )

A plot of this function is shown below.

From the graph, it is seen that I m a x ≈ 0.3149 I_{max} \approx 0.3149 and I m i n = − 0.192 I_{min} = -0.192 , so that their product is ≈ − 0.06047 \approx \boxed{-0.06047}

2 Helpful 2 Interesting 4 Brilliant 0 Confused

Great solution using Laplace transforms! I thought that was the way to go but didn't know how. After first mentioning partial fractions, you are missing a fraction line in the next equation for I ( s ) I(s) .

I ( s ) = s ( 1 + s ) ( s 2 + s + 1 ) = A s + 1 + B s + C s 2 + s + 1 I(s) = \dfrac{s}{(1+s)(s^2+s+1)} = \dfrac{A}{s+1} + \dfrac{Bs+C}{s^2+s+1}

Matthew Feig - 6 months, 2 weeks ago

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Thanks for correcting.

Hosam Hajjir - 6 months, 2 weeks ago

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Just curious: How would you adjust this approach if the initial conditions were different (the current or charge were nonzero at time zero)? I cannot tell where this information gets used in your solution.

Matthew Feig - 6 months, 2 weeks ago

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@Matthew Feig – Since the voltage across the inductor is v L ( t ) = L d i L ( t ) d t v_L(t) = L \dfrac{d i_L(t) }{dt} , then V L ( s ) = s L I ( s ) − L i ( 0 ) V_L(s) = s L I(s) - L i(0) , so that the loop equation becomes V S ( s ) = R I ( s ) + s L I ( s ) − L i ( 0 ) + 1 C s I ( s ) V_S(s) = R I(s) + s L I(s) - L i(0) + \dfrac{1}{C s} I(s) .

Hosam Hajjir - 6 months, 2 weeks ago

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@Hosam Hajjir – Thanks. I'll have to think that through!

Matthew Feig - 6 months, 2 weeks ago

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@Matthew Feig – I've corrected the equation in my previous comment to include a non-unity C C .

Hosam Hajjir - 6 months, 2 weeks ago

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