Impulse function applied to RLC circuit

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SUMMARY

The discussion centers on the application of the impulse function to a series RLC circuit, specifically analyzing the current i(t) at time t=0+ when a voltage source Vo*delta(t) is applied. The established conclusion is that the current Io at t=0+ is equal to Vo/L, where L is the inductance of the circuit. The confusion arises from the behavior of the inductor and capacitor during the instantaneous voltage change, with the inductor acting as a short circuit and the capacitor as an open circuit. The correct approach involves writing the differential equation and integrating it across the interval from 0- to 0+ to derive Io.

PREREQUISITES
  • Understanding of RLC circuit theory
  • Familiarity with impulse functions and delta functions
  • Knowledge of differential equations
  • Basic principles of circuit analysis
NEXT STEPS
  • Study the application of the impulse function in circuit analysis
  • Learn how to derive current and voltage equations in RLC circuits
  • Explore the integration of differential equations in electrical engineering
  • Review the concepts of short circuits and open circuits in reactive components
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Electrical engineering students, circuit designers, and anyone studying transient responses in RLC circuits will benefit from this discussion.

marc.orr
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Hello, I've been reviewing some material from my undergraduate studies. In the book "Signals and Systems" by Haykin in the first chapter he gives an example of a voltage source with the value of Vo*delta(t) applied to a series RLC circuit.

Vo is some arbitrary constant and delta(t) is the impulse function. The question is what is the value of the current i(t) at time t=0+ where 0+ is at 0 looking from the more positive side of time.

He says the answer is (without any sort of derivation) Io = Vo/L. How is this possible? I figure that as there is an instantaneous change in voltage, the inductor will look like a short circuit and the capacitor will look like an open circuit. So how can there be any current flow?

Thankyou in advance.

Marc
 
Last edited:
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marc.orr said:
I figure that as there is an instantaneous change in voltage, the inductor will look like a short circuit and the capacitor will look like an open circuit.
You have this reversed.
 
If you write the differential equation and integrate it between 0- and 0+ you can find Io.
 

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