How Do You Calculate iC(t) in an RLC Circuit and Plot Its Waveform?

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Discussion Overview

The discussion focuses on calculating the current iC(t) in an RLC circuit and plotting its waveform for t >= 0. Participants explore the circuit configuration, apply relevant equations, and discuss their approaches to solving the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about whether the circuit is a parallel or series RLC circuit after a switch moves, suggesting it may be series due to the current source.
  • Another participant proposes using Thevenin's theorem to simplify the circuit into a series configuration for t > 0.
  • A different participant describes using source transformation to analyze the circuit, calculating the initial current iL(0) and voltage Vc(0) across resistors.
  • This participant also derives the damping factor and natural frequency, concluding that the system is overdamped and calculates the coefficients A1 and A2 for the voltage across the capacitor.
  • One participant confirms that the calculations presented are correct so far.

Areas of Agreement / Disagreement

Participants generally agree on the calculations presented, with one participant confirming the correctness of the previous work. However, there is some uncertainty regarding the initial circuit configuration and the application of Thevenin's theorem.

Contextual Notes

Some assumptions regarding circuit behavior and initial conditions are present, but they remain unresolved. The discussion does not clarify all mathematical steps or dependencies on specific definitions.

Who May Find This Useful

Students studying circuit analysis, particularly those working on RLC circuits and their transient responses, may find this discussion beneficial.

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Homework Statement


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find iC(t) in the circuit and plot its waveform for t>=0.

Homework Equations


i(t)=i(infinity)+e^(-at)[D1cos(wdt)+D2sin(wdt)
D1= i(0)-i(infinity)
D2=i'(0)+a[i(0)-i(infinity)]

aux.:
w0=(LC)^(-1/2)
wd=sqrt(W0^2 - a^2)
s1= -a+sqrt(a^2 - w0^2)
s2= -a-sqrt(a^2 - w0^2)

The Attempt at a Solution


just a bit confused would this be a parallel RLC after all? doesn't seem like it, cause after switch moves it seems to be in series, except for current srouce. then I would calculate R(eq) by adding those two resistance? 0.1+0.52=0.62ohm?? i also get iL(0-)=iL(0)=4mA and iC(0)=0 because no current through open circuit?
 
Last edited:
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You could change the 4mA current source and its parallel 0.1 Ω resistor into its Thevenin equivalent. Then you'd definitely have a series circuit for t > 0.
 
i used source transformation and brought the 0.1ohm on top, and got an 0.4mV source. then i divided by the R(eq) of the two 0.1ohm resistors in series to get iL(0)=iL(0-)=2mA. then Vc(0) would be the voltage across the 0.1ohm resistors, 0.2mV. since iC(0)=iL(0) then using equaiton I=C(dv/dt) i get v'C(0)= iL(0)/C= 2mA*1.8=3.6mV. also, vC(infinity)=V(s)=0.4mV.

a (for series)= R/2L= 0.62ohm/(2*0.05)=6.2 np/s
w0 = LC^(-1/2)= 6 rad/s

since a>w0 it would be overdamped. so using the formula i listed i get s1=−4.64 and s2=−7.76
A1= (v'(0)-s2[v(0)-v(infnity)])/ s1-s2 = (3.6mV-(-7.76)(0.2mV-0.4mV))/3.12= 0.66mV
A2= -[(v'(0)-s1[v(0)-v(infnity)])/ s1-s2] = (3.6mV-(-4.64)(0.2mV-0.4mV))/3.12= -0.86mVvC(t) =vC(infnity) +A1*e^(s1t)+A2*e^(s2t)
vC(t) = 0.4mV+0.66mV*e^(-4.64t) - 0.86mV*e^(-7.76t)

so far i did this right?
 
Last edited:
Yup. Looks good so far.
 

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