Solve Overdamped RLC Circuit | i_R(t)

  • Context: Engineering 
  • Thread starter Thread starter jesuslovesu
  • Start date Start date
  • Tags Tags
    Circuit Rlc Rlc circuit
Click For Summary
SUMMARY

The discussion centers on solving for the current i_R(t) in an overdamped parallel RLC circuit with specified values: R = 20 mΩ, L = 2 mH, and C = 50 mF. The initial conditions are v(0+) = 0 and i(0-) = 2 mA. The correct initial current i_R(0^+) is confirmed to be zero, despite PSPICE simulations suggesting otherwise. The equations governing the circuit behavior are α = 500 Hz, ω₀ = 100 Hz, and i_R(t) = -2e^{-10.10t} + 2e^{-989.9t} mA.

PREREQUISITES
  • Understanding of parallel RLC circuit dynamics
  • Familiarity with differential equations in electrical engineering
  • Knowledge of PSPICE simulation software
  • Concept of overdamped response in RLC circuits
NEXT STEPS
  • Study the derivation of the differential equation for parallel RLC circuits
  • Learn how to analyze circuit responses using PSPICE
  • Explore the implications of overdamping in RLC circuits
  • Investigate the relationship between voltage and current in capacitors and inductors
USEFUL FOR

Electrical engineers, students studying circuit theory, and anyone involved in analyzing RLC circuit behavior will benefit from this discussion.

jesuslovesu
Messages
185
Reaction score
0
[SOLVED] Overdamped RLC circuit

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?

Homework Equations



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

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:
Physics news on Phys.org
jesuslovesu said:

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.

Homework Equations



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

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]
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 28 ·
Replies
28
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K