A 7.32 µF capacitor is charged with a 95.8 V battery. Now the battery is disconnected, and the capacitor is connected in series to, a 8.68 mH resistanceless coil, a 19.4 Q resistor and an open switch. At t = 0, the switch is closed. Find the energy, in µJ, stored in the capacitor at t = 0.708 ms. HINT : Is this circuit over, under or critically damped?

Delmar's Standard Textbook Of Electricity
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Chapter21: Resistive-capacitive Series Circuits
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A 7.32 µF capacitor is charged with a 95.8 V battery. Now the battery is disconnected, and the capacitor is connected in series to, a 8.68 mH resistanceless coil, a
19.4 Q resistor and an open switch. At t = 0, the switch is closed. Find the energy, in µJ, stored in the capacitor at t = 0.708 ms.
HINT : Is this circuit over, under or critically damped?
Transcribed Image Text:A 7.32 µF capacitor is charged with a 95.8 V battery. Now the battery is disconnected, and the capacitor is connected in series to, a 8.68 mH resistanceless coil, a 19.4 Q resistor and an open switch. At t = 0, the switch is closed. Find the energy, in µJ, stored in the capacitor at t = 0.708 ms. HINT : Is this circuit over, under or critically damped?
Expert Solution
Step 1

Given: A charged capacitor is connected in series with a resistor and an inductor at t = 0 as shown in figure below:

Electrical Engineering homework question answer, step 1, image 1

Step 2

Order of the circuit is two, with characteristic equation roots given as s1,2=-α±α2-ω02.

Here,

α=R2L,ω0=1LC.

Substituting the values,

α=R2L=19.42×8.68×10-3=1117.51,ω0=1LC=18.68×10-3×7.32×10-6=396.72.

α>ω0, we have the overdamped case.

The voltage across capacitor after time t = 0, has a form

vct=Ae-s1t+Be-s2t.

 

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