Second Order Resistor Capacitor Inductor Parallel (RCL) Circuit

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Parallel RLC Circuit - Energy Stored In Inductor


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Parallel RLC Circuit - Energy Stored In Inductor
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Second Order Resistor Capacitor Inductor Parallel Circuit
Figure 5.58 is the RCL circuit for the given circuit elements:
voltage source, Vs1 = 15 V; Rs1 = 130 kΩ;
inductance, L = 17 mH; R1 = 1.1 kΩ
Capacitance, C = 0.35 μF; R2 = 700 Ω Rs2 = 290 kΩ;
voltage source Vs2 = 9 V;
Assume DC steady state conditions exist for t < 0 and as t approaches ∞
If switch is open at t = 0; Determine:
(a) Energy stored in inductor at t = 0+, just after the switch S is open.
(b) Energy stored in inductor as t → ∞

The strings:
(a) and (b): S7P3A32 (Force - Push)
The math:
Second Order Resistor Capacitor Inductor Parallel Series Circuit
Pj Problem of Interest is of type force. Work and energy problems are force problems.
(a) DC steady state conditions exist for t < 0
So, inductor voltage, vL(0-) = 0.
And capacitor current, iC(0-) = 0.
So, iL(0-) = is1(0-) + is2(0-)
So, iL(0-) = vs1/Rs1 + vs2/Rs2
So, iL(0-) = 15/130 + 9/290 = 0.11538 + 0.03103 = 146.4 mA.
iL(0-) = iL(0+) (for continuity).
So, iL(0+) = 146.4 mA

Energy stored in inductor = (1/2)L(iL)2

So, Energy in inductor = (1/2) x (17 x 10-3) x (146.4 x 10-3)2 = 0.182 mJ.

(b)steady state conditions exist for t → ∞
At this time switch S has been open for a long time.
So, iL(∞) = vs2/Rs2 = 9/290 = 0.03103 = 31.03 mA.
So, Energy in inductor = (1/2) x (17 x 10-3) x (31.03 x 10-3)2 = 0 mJ.

Energy stored in a capacitor = (1/2)C(ic)2

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