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How do I solve a Series RLC circuit?
- Thread starter wiz0r
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In summary: I'll come back and give a more detailed and comprehensive review after I've had a chance to play with the software more.
- #2
technician
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I think you need to know the frequency of the supply to get started... do you have it?
- #3
wiz0r
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technician said:
The source is just a DC source... so, there's no frequency.
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technician
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my apologies... I thought it was an AC circuit
- #5
gneill
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Hi wiz0r,
Presumably the initial conditions that you're looking for is the state of the circuit at the time t=0+ when the switch has just opened.
Immediately before the switch opens, and assuming that the switch had been closed for "a long time", then your statement that the capacitor will look like an open circuit is correct -- no current will flow into or out of it. Similarly, the inductor will look like a short circuit.
Can you determine what voltage should be across the capacitor at t=0-?
Presumably the initial conditions that you're looking for is the state of the circuit at the time t=0+ when the switch has just opened.
Immediately before the switch opens, and assuming that the switch had been closed for "a long time", then your statement that the capacitor will look like an open circuit is correct -- no current will flow into or out of it. Similarly, the inductor will look like a short circuit.
Can you determine what voltage should be across the capacitor at t=0-?
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wiz0r
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- #7
gneill
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wiz0r said:
That looks fine. So when the switch opens the capacitor will have that voltage on it as an initial condition. And the current will still be zero because of the inductor.
To find i(t) you'll have to either write and solve the differential equation for the new circuit configuration, or use a "canned" solution for the RLC circuit.
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wiz0r
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- #9
gneill
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wiz0r said:
It's looking good!
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wiz0r
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gneill
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The equation should yield current values, and Vc(0) is a voltage.
At time t = 0 the current is zero, so K1 + K2 = 0. That is, K1 = -K2. You'll need to find another way to solve for the values. Perhaps if you could determine the initial rate of change of the current...
At time t = 0 the current is zero, so K1 + K2 = 0. That is, K1 = -K2. You'll need to find another way to solve for the values. Perhaps if you could determine the initial rate of change of the current...
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wiz0r
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- #13
gneill
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Rightwiz0r said:
s1K1 + s2K2 won't be zero. It'll equal the initial rate of change of the current. Go back to the circuit diagram and see if you can't determine what that might be. You've an inductor that has suddenly had a voltage Vc impressed upon it... how will it react?
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wiz0r
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gneill
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wiz0r said:
The equation of interest for the inductor is its defining equation: V = L dI/dt. In this case there's an impressed voltage Vc across inductor L=2.5H. What's dI/dt?
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wiz0r
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- #17
gneill
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wiz0r said:
Right. At t=0 dI/dt = Vc/L = 10.286 A/s. So, given K1 = -K2,...
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wiz0r
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gneill
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wiz0r said:
Looks good! Bravo!
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gneill
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wiz0r
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gneill said:Looks okay. I think if you made the maximum timestep a bit smaller you might resolve the peak a bit better (eliminate the "flat bottom".
Here's the plot of the mathematical version:
I've negated the current values ( -I(t) ) in order to match your Spice model current direction.
Perfect... thank you so much... I owe you big time... If I can repay you with something (except money.. I'm poor!) let me know! :)
- #23
gneill
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wiz0r said:
You're quite welcome. Just go get an "A" in your course!
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wiz0r
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gneill said:
I will definitely try my best. :)
Thank you so much. Bye bye for now. :)
Edwin
FAQ: How do I solve a Series RLC circuit?
What is a Series RLC circuit?
A Series RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C) connected in series. It is also known as a second-order circuit and is commonly used in electronic filters and oscillators.
What is the purpose of a Series RLC circuit?
The purpose of a Series RLC circuit is to control the flow of current in an electric circuit. The resistor, inductor, and capacitor work together to create a filter that can either block or pass certain frequencies, making it useful for various applications such as signal processing and frequency selection.
How does a Series RLC circuit behave in different frequencies?
In a Series RLC circuit, the behavior can vary depending on the frequency of the input signal. At low frequencies, the inductor has a higher impedance, and the circuit behaves like a high-pass filter. At high frequencies, the capacitor has a lower impedance, and the circuit behaves like a low-pass filter. At a certain resonant frequency, the impedance of the inductor and capacitor cancel out, resulting in a high impedance and a sharp response.
How do I calculate the resonant frequency of a Series RLC circuit?
The resonant frequency of a Series RLC circuit can be calculated using the formula: fr = 1/(2π√(LC)), where fr is the resonant frequency, L is the inductance in henries, and C is the capacitance in farads.
What is the difference between a Series RLC circuit and a parallel RLC circuit?
A Series RLC circuit has the components connected in series, whereas a parallel RLC circuit has the components connected in parallel. This means that the voltage across each component in a Series RLC circuit is the same, while the current through each component in a parallel RLC circuit is the same. Additionally, the resonant frequency of a parallel RLC circuit is calculated using the formula: fr = 1/(2π√(LC)), where fr is the resonant frequency, L is the inductance in henries, and C is the capacitance in farads.