Finding Current in a Parallel RLC Circuit

In summary, the homework statement is looking for an equation for the current through an inductor in a circuit with no resistor in parallel. Using the Kirchhoff's Voltage Law and simple algebra, the current equation is found to be 66.67 sin(480t) e-640t.
  • #1
Drakkith
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Homework Statement


Find io for t≥0.
Express your answer in terms of t, where t is in milliseconds.

There is no energy stored in the circuit in (Figure 1) when the switch is closed at t = 0.
8.34.jpg


Homework Equations


##x(t)=x_f+[B_1cos(ω_dt)+B_2sin(ω_dt)]e^{-αt}##

The Attempt at a Solution


Let ##i_R## be the current through the resistor, ##i_c## be the current through the capacitor, and ##V_c## be the voltage across the capacitor.
At ##t=0##:
No energy is stored in the circuit, so ##V_c(t<0)=0##, ##i_o(t<0)=0##, and since they are in parallel, ##V_o(t<0) = 0##.
Once the switch is closed, all of the current initially starts flowing through the capacitor. Since ##V_c(0^+)=V_c(0)=V_c(0^-)##, all of the voltage is across the resistor and ##i_R(0^+)=\frac{25}{125}=0.2A##.

To find the equation for ##i_o##, I believe I need ##i_f=i_o(∞)##. However, since there isn't a resistor in parallel with the LC portion of the circuit, there is nothing to absorb the energy of the oscillating current and it appears the current through the inductor oscillates forever. So what do I put for ##i_f##?
 

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  • #2
As time goes to infinity, the capacitor is an open circuit and the inductor is a short circuit, so Vo-->0 again. The 125 Ohm resistor provides the damping for the overall circuit.

I would write the KCL for the top right node and solve the DE...
Drakkith said:
x(t)=xf+[B1cos(ωdt)+B2sin(ωdt)]e−αtx(t)=x_f+[B_1cos(ω_dt)+B_2sin(ω_dt)]e^{-αt}
What's x(t)?
 
  • #3
berkeman said:
As time goes to infinity, the capacitor is an open circuit and the inductor is a short circuit, so Vo-->0 again. The 125 Ohm resistor provides the damping for the overall circuit.

Can you elaborate? I don't understand why the resistor would provide damping to the circuit since only the inductor and capacitor are in parallel.

berkeman said:
What's x(t)?

Insert V or i instead of x's there. It's just the general form of the formula for underdamped RLC circuits.

berkeman said:
I would write the KCL for the top right node and solve the DE...

I'll give it a shot.
 
  • #4
KCL at the top node:
##i_o+i_c=i_R##
##i_o=i_R-i_c##
To find the current, I turned everything into voltage:
##\frac{\int V_L}{L}=\frac{25-V_L}{125}-C\frac{dV_c}{dt}##
Taking the derivative gives:
##\frac{Cd^2}{dt^2}-\frac{dV}{125dt}-\frac{V}{L}=0##
Dividing by C and solving the differential equation yields:
##V_c=[B_1cos(480t)+B_2sin(480t)]e^{-640t}##

To find ##B_1## and ##B_2##:
##V(0)=V(∞)+B_1##
##0=0+B_1##
##B_1=0##

##\frac{dV(0)}{dt}=-640B_1+480B_2##
##\frac{0.2}{6.25*10^{-6}}=480B_2##
##B_2=66.67##

So ##V(t)=66.67sin(480t)e^{-640t}##

Plugging that back into the current equation ##i_o=i_R-i_c##:
##i_o=\frac{25-V}{125}-\frac{dV}{dt}=0.2+[-0.2cos(480t)-0.267sin(480t)]e^{-640t} A##

***Interestingly, this can be converted to an RLC circuit in parallel with a 0.2 Amp current sources instead of a voltage source.

Thanks @berkeman.
 
  • #5
Looks pretty good to me. Are you able to check the answer?
 
  • #6
Drakkith said:
Can you elaborate? I don't understand why the resistor would provide damping to the circuit since only the inductor and capacitor are in parallel.
The voltage source looks like an AC short circuit (current sources are AC open circuits), so that resistor actually is in parallel with the LC in this circuit.

Also, just stepping back and looking at the circuit at t=0- and t=infinity, you can see that Vo is going to be 0V. If it were an ideal parallel LC with no loss, then yes, after you hit it with some initial energy, it will ring forever. But in this case, the source is a DC voltage source with series resistor, so as t-->infinity, the DC voltage across the ideal inductor has to go to 0V (since its resistance is zero).

Does that help?
 
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  • #7
berkeman said:
The voltage source looks like an AC short circuit (current sources are AC open circuits), so that resistor actually is in parallel with the LC in this circuit.

Ah, okay. That makes sense.

berkeman said:
Also, just stepping back and looking at the circuit at t=0- and t=infinity, you can see that Vo is going to be 0V.

Once you recognize that the circuit is a parallel RLC circuit, sure. But I didn't realize it at the time (despite the thread title suggesting otherwise). I thought I had a parallel RLC circuit that was missing its parallel resistor and happened to be in series with another resistor.
 
  • #8
As a follow up to what berkman said, EVERY time I analyses a circuit I do a check at DC and inf current frequency. At DC just open all the capacitors and short all the inductors, at inf do the opposite.

After you start evaluating these circuits in the frequency domain rather than the time domain (which is much easier, algebra instead of calc) you will find that this analysis will allow you to check your own work 90% of the time.
 
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Related to Finding Current in a Parallel RLC Circuit

1. What is a parallel RLC circuit?

A parallel RLC circuit is an electrical circuit that contains three components - a resistor (R), an inductor (L), and a capacitor (C) - connected in parallel. This means that each component is connected to the same two points in the circuit, forming a branch. The equivalent resistance of a parallel RLC circuit is calculated differently than a series RLC circuit, and the behavior of the circuit can differ greatly.

2. How do I calculate the current in a parallel RLC circuit?

To calculate the current in a parallel RLC circuit, you can use the equation I = V/R, where I is the current in amperes, V is the voltage in volts, and R is the equivalent resistance of the circuit. The equivalent resistance can be calculated by using the formula 1/R = 1/RL + 1/RC + 1/RR, where RL is the resistance of the inductor, RC is the resistance of the capacitor, and RR is the resistance of the resistor. Once you have calculated the equivalent resistance, you can plug it into the first equation to find the current.

3. What is the phase difference in a parallel RLC circuit?

The phase difference in a parallel RLC circuit refers to the difference in the phase angles between the current and the voltage in the circuit. In a parallel RLC circuit, the voltage across each component is the same, but the current can differ in phase. The phase difference is determined by the ratio of the reactance (X) of the inductor and capacitor to the resistance (R). If XL > XC, the current will lead the voltage by a certain angle, and if XL < XC, the current will lag the voltage by a certain angle.

4. How does the frequency affect the current in a parallel RLC circuit?

The frequency of the current in a parallel RLC circuit can greatly affect the behavior of the circuit. At the resonant frequency, the reactance of the inductor and capacitor cancel each other out, resulting in a maximum current and minimum impedance. At other frequencies, the reactance of the inductor and capacitor may not cancel out, leading to a different current and impedance. This can also affect the phase difference between the current and voltage in the circuit.

5. What is the significance of a parallel RLC circuit in real-world applications?

Parallel RLC circuits are commonly used in electrical systems to filter out specific frequencies or to control the flow of current. They are commonly found in audio equipment, where they are used to filter out unwanted noise and improve sound quality. They are also used in power distribution systems to regulate the flow of electricity. Understanding the behavior of parallel RLC circuits is important for designing and troubleshooting these systems.

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