Second Order Resistor Capacitor Inductor (RCL) Series Circuits

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{4}A_{41} Base Sequence = 12735 String Sequence = 12735 - 4 - 41 **

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Second Order Resistor Capacitor Inductor (RCL) Series Circuits

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The circuit illustrated in figure 5.57 is a second order RCL circuit.

V_{s} = 12 V; C = 0.5 μF;

R_{1} = 31 kΩ; R_{2} = 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) S_{7}P_{4}A_{41} (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 S_{7}P_{4}A_{42} (Curved Motion).

(b) S_{7}P_{3}A_{32} (Force - Push).
**The math**:

(a) Pj Prblem of Interest is of type *motion* (Linear Motion).

At steady state:

Current through inductor = i_{L}(∞); voltage across inductor = v_{L}(∞) = 0;

Current through capacitor = i_{C}(∞) = 0;
voltage across capacitor = v_{C}(∞)

Applying Kirchoff Voltage Law (KVL) to loop acdf we have:

V - R_{1}i_{L}(∞) - R_{2}i_{L}(∞) = 0 ---------(1)

So, 12 - (31 x 10^{3})i_{L}(∞) - (22 x 10^{3})i_{L}(∞) = 0;

So, i_{L}(∞) = 12/(53 x 10^{3}) = 226 x 10^{-6}) = 226 μA.

(b) Pj Problem of Interest is of type *force* (Force - Push).

At steady state, i_{C}(∞) = 0;

So, v_{C}(∞) = v_{R2}(∞)

v_{R2}(∞) = i_{L}(∞)R_{2} = (22 x 10 ^{3}) x (226 x 10 ^{-6}) = 4.97 V.

So, v_{C}(∞) = 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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