Hey guys

I'm just wondering if i'm on the right track for the problem. I have started analysing the circuit, but I'm really unsure of my method of analysis.

The circuit switches from position "a" to position "b" at t=0
and I'm trying to fined i(t) for t>0

from what I understand, I need to determine i(o), which is for the circuit at t=0 (containing an inductor and 4A supply).

Then determine the α and wo for the circuit at t>0
where α = 1/2RC and wo = √(1/LC).
which will indicate the formula to use for the step response, to allow for determining i(t).


my calculated values so far-
by KCL
i(0) = 4A(6/2+8)
= 3A

at t>0
I used source transformation for the 4A supply (24V)
then added the total voltage in the loop (= +12V)

α = 1/(2 x 20Ω x 0.02F)
= 1.25

wo = √(1/ (2 x 0.02))
=5

therefor I understand I need to use
v(t) = (A1cos(wd)t + A2sin(wd)t)e^(-αt)

wd = √((wo^2) - (α^2))

= √(375/16)

... which isnt a very nice looking number

have I made a mistake somewhere? Is my whole understanding of the method in answering this question just plane wrong?

Thanks guys

ps. I'm really new to the whole posting on forums (this is the first forum I have ever used). So if I do anything the is annoying or if i'm doing anything i'm not supposed to be doing, plz tell me to I can correct myself.

 

my calculated values so far-
by KCL
i(0) = 4A(6/2+8)
= 3A

at t>0
I used source transformation for the 4A supply (24V)
then added the total voltage in the loop (= +12V)

α = 1/(2 x 20Ω x 0.02F)
= 1.25

wo = √(1/ (2 x 0.02))
=5

therefor I understand I need to use
v(t) = (A1cos(wd)t + A2sin(wd)t)e^(-αt)

wd = √((wo^2) - (α^2))

= √(375/16)

The damping factor you've calculated above seems to be the damping factor a parallel RLC circuit. We need the damping factor for a series RLC circuit.

Give it another whirl and see if you find the correct type of damping that the circuit is exhibiting.

 

ok I'v given it another shot, and it looks alot nicer.
but now I'm confused with the order I need to used the formulas

so I have for t>0
a = R/2L
=5

wo = √(1/ (2 x 0.02))
=5

therefore I use the critical dampening formula.
i(t) = (A2 + A1t)e^(-at)

now this is my confusion.
in my notes I have A1 = i(0), but for my question there is critical dampening when t>0.

so what/when "i' value would I use in the formula?
do I use it two times for i(0) and i(f)???

I'm crazy confused.

 

ok I'v given it another shot, and it looks alot nicer.
but now I'm confused with the order I need to used the formulas

so I have for t>0
a = R/2L
=5

wo = √(1/ (2 x 0.02))
=5

therefore I use the critical dampening formula.
i(t) = (A2 + A1t)e^(-at)

now this is my confusion.
in my notes I have A1 = i(0), but for my question there is critical dampening when t>0.

so what/when "i' value would I use in the formula?
do I use it two times for i(0) and i(f)???

I'm crazy confused.

You need to develope the ODE for this simple series circuit. Once you've done so you have formulate the characteristic equation which will provide you with two identical roots. (Multiplicity 2)

You know the form of the natural response will be,

\((A+Bt)e^{st}\)

where s is the root of your characteristic equation.

Add this to the forced response of your circuit and you will developed the complete response.

You must use the initial conditions of the circuit to solve for A and B.

 

sorry, but i'm not sure what ODE stands for. Is it the second order differential equation?

 

sorry, but i'm not sure what ODE stands for. Is it the second order differential equation?

Yes. You should be able to develope a Second Order Differential Equation. (ODE)

 

just a bit confused with determining the values used in the ODE.
am I using i(t) at 0 (i(0)) and the capacitor voltage at t = 0?
wouldn't the capacitor be zero at t = 0?
do I use R at t = 0 for the characteristic equation?

 

just a bit confused with determining the values used in the ODE.
am I using i(t) at 0 (i(0)) and the capacitor voltage at t = 0?
wouldn't the capacitor be zero at t = 0?
do I use R at t = 0 for the characteristic equation?

Why don't you show us what equation you are using for the 2nd Order ODE?

You have two unknowns to solve for so you would need two initial conditions, one being i(0) and the other Vc(0).

 

ok i'v started over, and this is where I'm up to.

by current divider in circuit formed by t<0
i(0) = 4(6/8) = i(0+)

voltage divider at t = 0 (my hand drawn cct)
Vc(0) = 12(14/30)
=4.8

by KVL
-12 +VL(0) + Vc(0) + 6i(0) + 14i(0)
VL(0) = 12 - Vc(0) - 4.8 - 6x3 - 14x3
=-52.8V

which looks wrong

a = R/2L = 20/(2x2H) = 5
wo = √(1/ (2 x 0.02))=5

a=wo
i(t) = (A + Bt)e^(-at)
where s1,2 = -a

and B = i(0) = 3A
A = di(0)/dt right???

I'm guessing there is alot wrong with what iv done so far.
if you could explain the things I got wrong so far, that would be amazing

 

voltage divider at t = 0 (my hand drawn cct)
Vc(0) = 12(14/30)
=4.8

To calculate Vc(0), you would want to consider whe the circuit has its switch on node a.

From here you can calculate Vc(0-) = Vc(0+). (There is no sudden change in voltage in a capacitor)

Now this part I myself am I little unsure about, so hopefully another member on this forum can verify my response.

First off the DC voltage source isn't providing any current because the other end of the circuit is not connected. (The switch is on a) With this there will be no voltage across the 14 ohm resistor, which leads me to think that all of the 12V is appearing across the capacitor.

I'm not sure whether or not that is correct, hopefully another member can chime in and clear things up.

If it is then I would assume that Vc(0-) = Vc(0+) = Vc(0) = 12V.

Hope this helps!

 

The initial capacitor voltage will be 0V [at t=0-]. Why? Prior to the switch transition there is no means by which the capacitor can acquire any charge. Since Vc cannot change instantaneously, the value of Vc at t=0+ will also be 0V.

 

The initial capacitor voltage will be 0V [at t=0-]. Why? Prior to the switch transition there is no means by which the capacitor can acquire any charge. Since Vc cannot change instantaneously, the value of Vc at t=0+ will also be 0V.

Thanks for clearing this up tnk!

 

ok i'v started over, and this is where I'm up to.

by current divider in circuit formed by t<0
i(0) = 4(6/8) = i(0+)

voltage divider at t = 0 (my hand drawn cct)
Vc(0) = 12(14/30)
=4.8

by KVL
-12 +VL(0) + Vc(0) + 6i(0) + 14i(0)
VL(0) = 12 - Vc(0) - 4.8 - 6x3 - 14x3
=-52.8V

which looks wrong

Effectively at t=0+ the circuit reduces to an inductor with an initial current of 3A, a capacitor with 0V, a total resistance of 20Ω in series with an equivalent source of 12V.

The same 3A initial inductor current flows in the 20Ω, producing a 60V drop.

With an equivalent source voltage of 12V and the 60V across the series resistance, the inductor must have a 48V drop of opposite polarity to the resistance drop. Keeping in mind the capacitor has zero voltage at t=0+.

So the general solution is

i(t)=(A+Bt)e^(-at)

At t=0, i(t)=3A which will give a value for the unknown A - given the 'Bt' term vanishes at t=0

At t=0, L*(di/dt)=-48V from which, after differentiation and substitution of values, you can find B, given the values A and a are already known.

 

sweet!
Great explanation!
so I got
A = 3
B = -33

so for t>0
i(t) = (3 - 33t)e^(-5t)
right?

 

sweet!
Great explanation!
so I got
A = 3
B = -33

so for t>0
i(t) = (3 - 33t)e^(-5t)
right?

B is incorrect.

Looks like you might have left out the 2H Inductance value in the solution.

The induced voltage drop in the inductance is e(t)=L*(di/dt)=-48

You seem to have solved (di/dt)=-48 at t=0

 

so what I did was
at t = 0
i=3A
3 = (A + 0)e^0
A=3

(di/dt) = -48 = (-5e^(-5t)) * (A +Bt) + e^(0)*B
are you saying it should be
di/dt = -24 = [(-5e^(-5t)) * (A +Bt) + e^(0)*B]/L

or am I missing a step?

 

are you saying it should be
di/dt = -24 = [(-5e^(-5t)) * (A +Bt) + e^(0)*B]/L

That's correct.

 

sweeeeeet!
thanks