# Series RLC circuit with source

• Engineering

## Homework Statement

R=50Ω
L=5H
C=1/125F
vs=200.[1+u(t)] V

Find i(t)

## The Attempt at a Solution

Write the KCL equation
-vs+Ri+L.i'+1/C.∫idt=0
Do the derivation to get rid of the integral, replace all possibly values, we got:
i''+10i'+25i=1/5.vs' (*)
solve A(s)=s2+10s+25=0, we got the natural response in(t)=A1+A2.e-5t

Replace the inductor with a short circuit, the capacitor with open circuit, so I have ip=0
and the initial condition i(0+)=i(0-)=0. (**)

So the solution should be i(t)=in+ip

I thought I was wrong at (**) and maybe at (*) cause the vs' would be zero and the solution would be zero as well.

## Answers and Replies

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rude man
Homework Helper
Gold Member
1. Replace the inductor with a short circuit, the capacitor with open circuit, so I have ip=0
and the initial condition i(0+)=i(0-)=0. (**)

So the solution should be i(t)=in+ip

I thought I was wrong at (**) and maybe at (*) cause the vs' would be zero and the solution would be zero as well.

i(0+) = i(0-) ≠ 0. Look at the definition of the input. What does vs=200.[1+u(t)] V mean?

rude man
Homework Helper
Gold Member
i''+10i'+25i=1/5.vs' (*)
solve A(s)=s2+10s+25=0, we got the natural response in(t)=A1+A2.e-5t

Replace the inductor with a short circuit, the capacitor with open circuit, so I have ip=0
and the initial condition i(0+)=i(0-)=0. (**)

So the solution should be i(t)=in+ip

I thought I was wrong at (**) and maybe at (*) cause the vs' would be zero and the solution would be zero as well.

Even if Vs = u(t) the current i(t) would not be zero! You are ignoring Vs!
You have to include Vs before trying to fit the initial conditions. Since this is a 2nd order ODE you know there have to be two initial conditions on i.