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Overdamped RLC circuit

  • Engineering
  • Thread starter jesuslovesu
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[SOLVED] Overdamped RLC circuit

1. Homework Statement

In a parallel RLC circuit determine [tex]i_R(t)[/tex].
R = 20 mohms
L = 2mH
C = 50 mF

v(0+) = 0 (capacitor)
i(0-) = 2mA (inductor)


My question is what is [tex]i_R(0^+)[/tex]? According to my final answer, it should be 0. However, went I graph it with PSPICE, it looks like it starts out somewhere at -186mA. I know that i_R can change instantaneously but the graph that pspice makes, makes it look like it will never be 0. I was under the impression that if the voltage of the capacitor is 0 then iR(t) will be 0 regardless of the current in the inductor. Is this correct?

2. Homework Equations

[tex]\alpha = 500 Hz[/tex]
[tex]\omega_0 = 100 Hz[/tex]
[tex]i_R(t) = Ae^{-10.10t} + Be^{-989.9t}[/tex]

3. The Attempt at a Solution

Using these two equations
A + B = 0
[tex]i_c(0+) + i_L(0+) + i_R(0+) = 0[/tex]
[tex]RC di_R/dt = -2 mA[/tex]
[tex]i_R(t) = -2e^{-10.10t} + 2e^{-989.9t} mA[/tex]
 
Last edited:

Answers and Replies

CEL
655
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1. Homework Statement

In a parallel RLC circuit determine [tex]i_R(t)[/tex].
R = 20 mohms
L = 2mH
C = 50 mF

v(0+) = 0 (capacitor)
i(0-) = 2mA (inductor)


My question is what is [tex]i_R(0^+)[/tex]? According to my final answer, it should be 0. However, went I graph it with PSPICE, it looks like it starts out somewhere at -186mA. I know that i_R can change instantaneously but the graph that pspice makes, makes it look like it will never be 0. I was under the impression that if the voltage of the capacitor is 0 then iR(t) will be 0 regardless of the current in the inductor. Is this correct?
Yes, you are correct. Since [tex]V_R(0^+) = V_C(0^+) = 0[/tex], [tex]i_R(0^+)[/tex] should also be zero.
2. Homework Equations

[tex]\alpha = 500 Hz[/tex]
[tex]\omega_0 = 100 Hz[/tex]
[tex]i_R(t) = Ae^{-10.10t} + Be^{-989.9t}[/tex]

3. The Attempt at a Solution

Using these two equations
A + B = 0
[tex]i_c(0+) + i_L(0+) + i_R(0+) = 0[/tex]
[tex]RC di_R/dt = -2 mA[/tex]
[tex]i_R(t) = -2e^{-10.10t} + 2e^{-989.9t} mA[/tex]
You should write your differential equation using [tex]i_L[/tex] as the independent variable.
You get [tex]i_L(t) = Ae^{-10.10t} + Be^{-989.9t}[/tex], with [tex]i_L(0) = 2mA[/tex] and [tex] \frac{di_L}{dt}(0) = 0[/tex]
 

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