How does a capacitor in an RLC circuit lose charge?

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SUMMARY

The discussion centers on the behavior of a capacitor in an RLC circuit, specifically a capacitor with a capacitance of 6.4 µF charged to 12V, which is then discharged through a resistor of 3 Ohms and an inductor of 2.0 x 10-3 H. The charge on the capacitor at the start of the discharge is equal to the initial charge, calculated as Q = C * V, resulting in 0.0768 C. The time for the charge amplitude to decrease to one-tenth of its initial value can be determined using the time constant formula for an RLC circuit, which involves both resistance and inductance.

PREREQUISITES
  • Understanding of RLC circuit dynamics
  • Knowledge of capacitor charge calculations
  • Familiarity with time constant in electrical circuits
  • Basic principles of inductance and resistance
NEXT STEPS
  • Study the mathematical derivation of the RLC circuit's charge decay formula
  • Learn about the impact of resistance on the time constant in RLC circuits
  • Explore simulations of RLC circuits using tools like LTspice
  • Investigate the role of damping in RLC circuits and its effects on charge loss
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in the transient response of RLC circuits will benefit from this discussion.

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1. First you fully charge a capacitor (C = 6.4 µF) with a 12V battery. Then you disconnect the battery and discharge the capacitor through a resistor (R = 3 Ohms) and an inductor (L = 2.0 x 10-3 H) in series.

2. What is the charge on the capacitor when you begin discharging at t= 0?

How long does it take for the charge amplitude to fall to one tenth of initial value?

At what time does the charge first fall to one tenth of its initial value?



The Attempt at a Solution


 
Physics news on Phys.org
Well what have you thought about doing? What dictates how fast a capacitor in an RLC (Resistor - Inductor - Capacitor) circuit will lose charge?
 

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