# Confusing RLC circuit..

Discussion in 'Homework Help' started by mesaman000, Jul 13, 2010.

1. ### mesaman000 Thread Starter New Member

Jun 29, 2010
16
0

So with this I honestly have no idea how to start.. Do I need to find the initial or final expressions to solve for this? Because if I were to find the final then the cap would act as an OC and the inductor would act as a Short circuit just leaving current division between those 2 resistors..

Is this the right way to approach this?

I know the formulas needed for energy of a inductor is E=.5*LI^2 and E=.5*CV^2

Can someone just help me get started with this question? I've dealt iwth RC and LC circuits but not with RLC:-/

2. ### mesaman000 Thread Starter New Member

Jun 29, 2010
16
0
And thet hing that I'm mainly confused about is that obviously you cant do this in teh frequency domain since youre not given enough info, so you have to assume its in the time domain

But should I analyze this circuit as t-> infinity? or at any instantaneous point? I just don't know what assumptions to make to start off with

I tried finding Req but I get stuck at the part hwere the resistor is in parallell with a resistor+inductor in series..

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
The problem says nothing about time, so assume the circuit has reached equilibrium. Find the voltage across the capacitor, calculate the energy stored there, and subtract from the total stored energy; the remainder must be the energy stored in the inductor. From that knowledge, you can calculate the value of L.

4. ### mesaman000 Thread Starter New Member

Jun 29, 2010
16
0
oh so is 226 mH right?

my initial thoughts were that all the energy, at steadystate/equilibrium would be stored into both the capacitor and inductor combined.. So thats why I was able to use the steady state expressiosn and treat the cap as an OC and inductor as a SC and then find the voltage across the cap to be 75 V... Does that sound right?

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
The voltage across the cap is 75 volts, but I get a different result for the inductor.

If you show the details of your calculation, I could help find your mistake (assuming I didn't make one!).

6. ### mesaman000 Thread Starter New Member

Jun 29, 2010
16
0
ok so 120mj= .5*cv^2 + .5 Li^2

120mj= .5*(20u)(75)^2 + .5Li^2
63.75=.5Li^2

since theres .75 A going down (using current division) through the right branch, then its

.5*(.75)^2 * L= 63.75

L=226 mH

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
If you have 75 volts across the capacitor, then you also have 75 volts across the series combination of L and 60Ω, for a current of 1.25 amps in the inductor.

Or, using current division, 2 amps * 100/(100+60) = 1.25 amps.