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Second Order Resistor Capacitor Inductor (RCL) Series Circuits


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Strings (SiPjAjk) = S7P4A41     Base Sequence = 12735     String Sequence = 12735 - 4 - 41



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Second Order Resistor Capacitor Inductor Series Circuit
The circuit illustrated in figure 5.57 is a second order RCL circuit.
Vs = 12 V; C = 0.5 μF;
R1 = 31 kΩ; R2 = 22 kΩ
L = 0.9 mH.
The switch is closed at t = 0 after having been open for an extended period of time.
Determine:
(a) The current through the inductor after the circuit has returned to steady state.
(b) The voltage across the capacitor after the circuit has returned to steady state.

The string:
(a) S7P4A41 (Linear Motion - Current Flowing Into Passive Elements). The motion of current flow in an electric circuit can be viewed as piece-wise-linear (current flow into a device) or looped (current flow in a circuit branch or the entire circuit). In a series connection, the piece-wise-linear current and looped current are the same. In a parallel connection, they are not. The looped current is stringed as S7P4A42 (Curved Motion).
(b) S7P3A32 (Force - Push).
The math:
Second Order Resistor Capacitor Inductor Series Circuit
(a) Pj Prblem of Interest is of type motion (Linear Motion).
At steady state:
Current through inductor = iL(∞); voltage across inductor = vL(∞) = 0;
Current through capacitor = iC(∞) = 0; voltage across capacitor = vC(∞)
Applying Kirchoff Voltage Law (KVL) to loop acdf we have:
V - R1iL(∞) - R2iL(∞) = 0 ---------(1)
So, 12 - (31 x 103)iL(∞) - (22 x 103)iL(∞) = 0;
So, iL(∞) = 12/(53 x 103) = 226 x 10-6) = 226 μA.
(b) Pj Problem of Interest is of type force (Force - Push).
At steady state, iC(∞) = 0;
So, vC(∞) = vR2(∞)
vR2(∞) = iL(∞)R2 = (22 x 10 3) x (226 x 10 -6) = 4.97 V.
So, vC(∞) = 4.97 V.

The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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