Solve Overdamped RLC Circuit | i_R(t)

[jesuslovesu]

[SOLVED] Overdamped RLC circuit

Homework Statement

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

v(0+) = 0 (capacitor)
i(0-) = 2mA (inductor)My question is what is i_R(0^+)? 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?

Homework Equations

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

The Attempt at a Solution

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

 

[CEL]

Homework Statement

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

v(0+) = 0 (capacitor)
i(0-) = 2mA (inductor)My question is what is i_R(0^+)? 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 V_R(0^+) = V_C(0^+) = 0, i_R(0^+) should also be zero.

Homework Equations

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

The Attempt at a Solution

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

You should write your differential equation using i_L as the independent variable.
You get i_L(t) = Ae^{-10.10t} + Be^{-989.9t}, with i_L(0) = 2mA and \frac{di_L}{dt}(0) = 0