# Overdamped RLC circuit

• Engineering
[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
$$i_c(0+) + i_L(0+) + i_R(0+) = 0$$
$$RC di_R/dt = -2 mA$$
$$i_R(t) = -2e^{-10.10t} + 2e^{-989.9t} mA$$

Last edited:

## 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 $$\frac{di_L}{dt}(0) = 0$$