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

 

[piercas]

First, it is important to note that the initial conditions given in the question (v(0+) = 0 and i(0-) = 2mA) are not consistent with an overdamped RLC circuit. In an overdamped circuit, the initial conditions for the capacitor (v(0+) = v(0-) = 0) and the inductor (i(0+) = i(0-) = 0) should be equal.

Assuming that the initial conditions for the capacitor and inductor are both 0, the solution for i_R(t) in an overdamped RLC circuit would be:

i_R(t) = Ae^{-10.10t} + Be^{-989.9t}

Where A and B are constants that can be determined using the initial conditions and the equations for the circuit.

Now, to address the discrepancy between your calculated solution and the graph from PSPICE, it is important to remember that PSPICE creates a numerical approximation of the solution, which may not be completely accurate. It is also possible that there could be a mistake in your calculations or equations that is causing the discrepancy. Double check your work and equations to ensure they are correct.

In addition, it is important to note that the solution for i_R(t) in an overdamped circuit will eventually approach 0, but it may take longer than a critically damped or underdamped circuit. This could explain why the graph from PSPICE appears to start at -186mA. However, it is important to remember that the solution for i_R(t) will eventually reach 0, even if it takes a longer time.

In conclusion, the solution for i_R(t) in an overdamped RLC circuit should approach 0, but it may take longer than other types of circuits. It is also important to double check your calculations and equations to ensure they are correct.