# Confusing RLC circuit..

#### mesaman000

Joined Jun 29, 2010
16

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:-/

#### mesaman000

Joined Jun 29, 2010
16
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..

#### The Electrician

Joined Oct 9, 2007
2,968
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.

#### mesaman000

Joined Jun 29, 2010
16
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?

#### The Electrician

Joined Oct 9, 2007
2,968
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!).

#### mesaman000

Joined Jun 29, 2010
16
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

#### The Electrician

Joined Oct 9, 2007
2,968
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
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.