# series-parallel RLC

Discussion in 'Homework Help' started by keltix, Jan 24, 2010.

1. ### keltix Thread Starter New Member

Jan 10, 2010
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0
i need help deriving the second order differential equation for this combination of series and parallel RLC:

i am supposed to end up with one equation...
i can use KVL and KCL to get two equations but i end up with more than 2 variables:

KVL - L*di(L)/dt+ i(R)*R= V in

KCL - i(vin) = i(C) + i(R)
(1/L)*integral of(v(t)) = C*dv(C)/dt + i(R)

Last edited: Jan 24, 2010
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Proceed as follows

iL(t)=iC(t)+iR(t) ---------------(1)

iL(t)=(1/L)∫(Vin-Vout)dt ------(2)

iC(t)=Cd(Vout)/dt -------------(3)

iR(t)=Vout/R ---------------------(4)

Substitute (2), (3) & (4) into (1)

Differentiate through once and you will have your second order equation with Vout as the dependent variable and Vin as the independent.

3. ### Luke Rodrick New Member

Jun 4, 2015
2
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Would the formula remain the same if a resistor was added in series between the Inductor and the voltage source?

4. ### WBahn Moderator

Mar 31, 2012
18,085
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If the formula remains the same that means that the resistor has not impact on the output. Does that make sense? Think of it in the extreme in which the resistor is so large in value that it is basically an open circuit. Do you expect the same output voltage as in the original circuit?

5. ### Luke Rodrick New Member

Jun 4, 2015
2
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That does make sense, I was thinking that the general solution would remain the same but now R would be replaced with R thevenin. Which would be the new R in parallel with the R to right. I meant what you said makes sense, not what I said in my original comment.

Last edited: Jun 4, 2015