Parallel RLC Circuit - Energy Stored In Inductor

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{3}A_{32} Base Sequence = 12735 String Sequence = 12735 - 3 - 32 **

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Parallel RLC Circuit - Energy Stored In Inductor

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Figure 5.58 is the RCL circuit for the given circuit elements:

voltage source, V_{s1} = 15 V; R_{s1} = 130 kΩ;

inductance, L = 17 mH; R_{1} = 1.1 kΩ

Capacitance, C = 0.35 μF; R_{2} = 700 Ω R_{s2} = 290 kΩ;

voltage source V_{s2} = 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): S_{7}P_{3}A_{32} (Force - Push)
**The math**:

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, v_{L}(0^{-}) = 0.

And capacitor current, i_{C}(0^{-}) = 0.

So, i_{L}(0^{-}) = i_{s1}(0^{-}) + i_{s2}(0^{-})

So, i_{L}(0^{-}) = v_{s1}/R_{s1} + v_{s2}/R_{s2}

So, i_{L}(0^{-}) = 15/130 + 9/290 = 0.11538 + 0.03103 = 146.4 mA.

i_{L}(0^{-}) = i_{L}(0^{+}) (for continuity).

So, i_{L}(0^{+}) = 146.4 mA

Energy stored in inductor = (1/2)L(i_{L})^{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, i_{L}(∞) = v_{s2}/R_{s2} = 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(i_{c})^{2}

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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