Derive time-dependent current for circuit

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SUMMARY

The discussion focuses on deriving the time-dependent current I(t) for an LC circuit involving an inductor (L) and a capacitor (C) after a switch is closed at t=0. Using Kirchhoff's Voltage Law (KVL), the voltage equations for the inductor (V_L = Ldi/dt) and capacitor (V_C = (1/C) ∫idt) are established. The resulting differential equation, Ld²i/dt² + (1/C)i = 0, is solved to yield the general solution I(t) = Acos(t/√(LC)) + Bsin(t/√(LC)), where constants A and B are determined by initial conditions.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Knowledge of differential equations
  • Familiarity with LC circuit dynamics
  • Basic calculus for integration and differentiation
NEXT STEPS
  • Research methods for solving second-order differential equations
  • Study the behavior of LC circuits under different initial conditions
  • Learn about the applications of oscillatory circuits in electronics
  • Explore the impact of resistance on LC circuit performance
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing oscillatory circuits will benefit from this discussion.

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Homework Statement


Derive an expression for the current I(t) as a function of time for the following circuit:

____ \____
|..... |
&..... -
& .....-
| ...|
___________

This circuit may be a bit unclear - the && is an inductor L, the - - is a capacitor C, and the top part is a switch that is open for t<0 and closed at t=0. And the ... are filler.

Homework Equations



I have no idea I am struggling incredibly.

The Attempt at a Solution



Help help please
 
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1. Use Kirchoff's Voltage Law (sum of voltages around the loop of the circuit):2. However, we need to determine the voltage across the inductor and capacitor.
Just do research, the voltage of the inductor is:

V_L = Ldi/dt

V_C = (1/C) integral (idt)3. Using KVL: Ldi/dt + (1/C)integral(idt) = 04. We need to turn this one to differential equations. So we differentiate term by term

Ld^2i/dt^2 + (1/C)i = 05. Use methods of differential equation to solve for i with respect to t.

Lm^2 + (1/C) = 0

m^2 + 1/(L*C) = 0; m = +- j/(L*C)

I(t) = Acos(t/(LC)) + Bsin(t/(LC))6. The constants A and B can be derived using initial conditions.

See here for more details:

http://en.wikipedia.org/wiki/LC_circuit
 

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