RLC initial conditions

by axsvl77   Last Updated October 10, 2019 02:25 AM - source

I am working on the following series RLC circuit: enter image description here

Let the circuit have non-zero initial conditions \$I(0^+)= 6\text{A}\$ and a voltage across the capacitor of \$V(0^+)=-12\text{V}\$.

What are \$\frac{dV(0^+)}{dt}\$ and \$\frac{dI(0^+)}{dt}\$?

My solution is different from the text example, and I want to know why.

$$I=\frac{dv}{dt} ~~\rightarrow ~~ \frac{dV(0^+)}{dt}=\frac{1}{C}I(0^+) = \frac{6}{C} $$

and

$$ v = L\frac{dI}{dt} ~~ \rightarrow ~~ \frac{dI(0^+)}{dt} = \frac{1}{L}V(0^+) =- \frac{12}{L}$$

The text, on the other hand, shows: $$\frac{dI(0^+)}{dt} = -\frac{1}{L}\Big[RI(0^+) + V(0^+)\Big]$$

What is giong on here? What am I doing wrong?



Answers 1


What is giong on here? What am I doing wrong?

The first thing I see is here,

$$ v = \frac{dI}{dt} ~~ \rightarrow ~~ \frac{dI(0^+)}{dt} = \frac{1}{L}V(0^+) =- \frac{12}{L}$$

You assumed the voltage across the inductor is equal to the voltage across the capacitor. But given 6 A flowing around the circuit, the voltage across the resistor is not zero and is probably not small enough to be neglected.

You should use KVL, considering the drop across both the capacitor and the resistor, to find the voltage across the inductor. Which will give you the rate of change in the current.

The Photon
The Photon
October 10, 2019 02:19 AM

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