# RLC, when in series ? when in parallel ?

Discussion in 'Homework Help' started by Hitman6267, May 29, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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0

In the solution of this problem they wrote alpha= R/2L. That formula applies when its a series RLC. What elements of the circuit are we talking about when we're saying series/parallel. I used to think it was the inductance and the capacitor. Does opening the switch making the inductance capacitor resistor a separate circuit has anything to do with it ?

Here's how I think if i'm asked to find i(t) -current through the inductance - for t > or = 0

I get i(0) so I get a relationship like this A1+A2 =0

then I derive the general equation of the case.
=> di/dt = ...
then I think that di/dt isn't "anything" so I remember that the voltage of a inductance is V= L di/dt. so I multiply both sides of the newly derived equation to get V= ...
In this circuit V(0) would be 0 because the inductance became a wire.

Is this a wrong way to think ? It's never working for me.

Back to the solution I posted. How did they find di(0)/dt ?

Last edited: May 29, 2010
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Yes - this is a series circuit. There is only one path through which current may flow around the remaining RH loop when the switch opens - hence no parallel paths exist.

If you consider the initial circuit conditions - starting with the general DE for vo(t) in terms of the LH side of the loop

$v_{o}(t)=Ri_{o}(t)+L\frac{di_{o}(t)}{dt}$

Hence

$\frac{di_{o}(t)}{dt}=\frac{1}{L}(v_{o}(t)-Ri_{o}(t))$

At t=0

$v_{o}(0)=75V$

$i_{o}(0)=15mA$

$\frac{di_{o}(0)}{dt}=\frac{1}{L}(v_{o}(0)-Ri_{o}(0))$

$\frac{di_{o}(0)}{dt}=\frac{1}{1}(75-5000*15e^{-3})=0$

Apr 6, 2010
82
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4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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$15e^{-3}$

is meant to be scientific format notation - for the 15mA value. I'm surprised you didn't realise that.

perhaps I should have either written more correctly [by convention]

$15E^{-3}$

or

$15*10^{-3}$

I tend to use notations interchangeably & somewhat ambiguously perhaps.

In any case the solution I gave is correct if you allow my intended meaning.

Hitman6267 likes this.
5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
ok That makes sense. Thank you.