Finding vC(t) for a Critical Damping RLC Circuit

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SUMMARY

The discussion centers on finding the voltage across the capacitor, vC(t), in a critically damped RLC circuit with given initial conditions: a capacitor voltage of 15 V and an inductor current of 6 mA, with a resistance of 1250 ohms. The differential equation yields two roots of -5000, leading to the general solution v(t) = C_1*e^(-5000t) + C_2*t*e^(-5000t). The constants C_1 and C_2 are derived from initial conditions, with C_1 equating to 15 V. However, a discrepancy arises in calculating C_2, which should be 56250 instead of 93750, indicating a miscalculation in the application of the initial conditions.

PREREQUISITES
  • Understanding of RLC circuit dynamics and critical damping.
  • Familiarity with differential equations and their solutions.
  • Knowledge of initial condition application in circuit analysis.
  • Proficiency in capacitor and inductor behavior in transient analysis.
NEXT STEPS
  • Review the derivation of solutions for second-order linear differential equations.
  • Study the application of initial conditions in RLC circuit analysis.
  • Learn about critically damped systems and their characteristics.
  • Explore the use of Laplace transforms in solving circuit equations.
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing RLC circuits and transient responses will benefit from this discussion.

MattHorbacz
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Homework Statement


Figure_P08.44.jpg

In the circuit in the following figure, the resistor is adjusted for critical damping. The initial capacitor voltage is 15 V, and the initial inductor current is 6 mA and R=1250 ohms

Find vC(t) for t≥0.
Express your answer in terms of t, where t is in milliseconds.

Homework Equations



i=c*dv/dt

The Attempt at a Solution


the only part of this problem i am not getting is solving for the constants with the given initial conditions... the 2 roots to the differential equation are -5000. so:
v(t)=C_1*e^(-5000t)+C_2*t*e^(-5000t)
and
dv/dt [at t=0+]=-5000*C_1+C_2=i/c=(6*10^-3)A/(320*10^-9 F)=18750 volts/s
v(0)=C_1=15V
so C_2 -5000*15=18750...C_2=93750
but apparently C_2 is actually equal to 56250...If the equation was C_2-5000*15=-18750, then i would get 56250...where am i going wrong?
 
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As the charge flows away from the capacitor, I=-dQ/dt= - C dV/dt.

ehild
 

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