find the RLC equation

hello
as you see this circuit i want to find Ic(0+).how can i do that?

where does this Ic(0+) flow ? Is it the capacitor ? It's not marked anywhere.
Anyhow, Voltage across capacitor will be 8 volts ( 1A x 8 ohms), assuming steady state before switch closure. And current thru' capacitor at t = 0- will be zero.

At t=0+, cap discharges thru' inductor. You know how to work out what happens there.

cheers!

Hello,

I remember a similar problem with your capacitor booster circuit. Are you familiar with solving linear differential equations?

Hello
I'm reading the circuit analysis.i know writing of differential equations but not very familiar to solve them with mathematical ways.
could you tell me the best reference to know them?what you suggest?

in the above circuit i want to know how can find the capacitor current at 0+

baby_1 said:

Hello
I'm reading the circuit analysis.i know writing of differential equations but not very familiar to solve them with mathematical ways.
could you tell me the best reference to know them?what you suggest?

Click to expand...

Exercises, exercises and after... more exercises.

in the above circuit i want to know how can find the capacitor current at 0+

Click to expand...

At 0- the system is in steady state ==> Ic = 0 and V = IR = 8V

At 0+ the switch is closed, but the current at L is still 0 ( due the current continuity because Vl = L dIl/dt and Vl is a step), then Ic(0+) = 0.

baby_1 said:

I'm reading the circuit analysis.i know writing of differential equations but not very familiar to solve them with mathematical ways.

Click to expand...

You could use a circuit simulator and let your computer do the maths for you.

Here's the result with the switch closing at T = 1 second.

Thanks
could you tell me a book that i can read more and find many problems in it?

Hello,

When you put: damped oscillation LC differential equations in a search engine, you will find useful info. If you still find difficulties, you can of course post them. the result containing "Unit 5 Electrical Damped Oscillation" is good as it doesn't require laplace transform.

The CORRECT answer is actually --

Inductor current at T=0⁺ Infinity + 1

The infinite comes from the capacitor, and the +1 comes from the current source.

Capacitor current at t=0⁺ is Infinite

Note that an ideal inductor has zero resistance. At t=0+ there is NO magnetic field present.
Naturally this situation only lasts for delta --> 0 time period.

rohitkhanna,

Capacitor current at t=0+ is Infinite

Note that an ideal inductor has zero resistance. At t=0+ there is NO magnetic field present.
Naturally this situation only lasts for delta --> 0 time period.

Click to expand...

I disagree. The inductance will keep the current at a finite level regardless of the resistance. See analysis below.

Ratch

LoL !! :smile:
very pretty equations Ratch. I'm sure they are correct.

HOWEVER, we are NOT talking about steady-state analysis.
We are ONLY talking about the infinitisemal time immediately after closure of the "perfect" switch.
Thats what t=0⁺ means.

The capacitor current is never infinite.

An ideal inductor has zero DC resistance, but it has inductance. When a voltage is applied, the current increases from zero at a rate that depends on the inductance and the voltage.

Even with a perfect inductor, it would take an infinite amount of time for the current to rise to infinity.

rohitkhanna,

HOWEVER, we are NOT talking about steady-state analysis.
We are ONLY talking about the infinitisemal time immediately after closure of the "perfect" switch.
Thats what t=0+ means.

Click to expand...

The Laplace transform gives the total response. As you can see, there is no high current spike, just a nondecaying sinusoidal response. In order to obtain a high current spike, you would have to have a perfect cap shorted out by a superconductor wire.

Ratch

godfreyl said:

The capacitor current is never infinite.

An ideal inductor has zero DC resistance, but it has inductance. When a voltage is applied, the current increases from zero at a rate that depends on the inductance and the voltage.

Even with a perfect inductor, it would take an infinite amount of time for the current to rise to infinity.

Click to expand...

(sigh)
Ok fine.

Lets take the converse case of a CAPACITOR - an ideal one with zero initial charge, no resistance, no inductance - being instantantly connected to an ideal voltage source source (ideal here means - zero output resistance, infinite current capability).

What do you think will be the current at the instant of connection ? I claim it will be infinite and the cap will charge up to the voltage of the source instantly. Why would it not ? What is there to stop the current going to infinity ?

Similarly, when you connect an ideal Cap to an ideal Inductor 'suddenly', what exists at THAT INSTANT OF TIME to stop the flow of current ?

---------- Post added at 21:26 ---------- Previous post was at 21:21 ----------

Ratch said:

rohitkhanna,




The Laplace transform gives the total response. As you can see, there is no high current spike, just a nondecaying sinusoidal response. In order to obtain a high current spike, you would have to have a perfect cap shorted out by a superconductor wire.

Ratch

Click to expand...

Really ? Laplace does that ? I always thought it worked on linear integrals & differentials, not on discontinous functions.
So whats the Laplace transform of the Dirac delta then ? I need to know this for a project I'm working on.....

Secondly - regarding the perfect cap etc etc comment. Is that not what the question was all about ? A perfect cap of 0.4F and a perfect Inductor ( i.e. ZERO resistance - never mind the superconducting point, & zero capacitance) of 1H etc ??

rohitkhanna,

Lets take the converse case of a CAPACITOR - an ideal one with zero initial charge, no resistance, no inductance - being instantantly connected to an ideal voltage source source (ideal here means - zero output resistance, infinite current capability).
What do you think will be the current at the instant of connection ? I claim it will be infinite and the cap will charge up to the voltage of the source instantly. Why would it not ? What is there to stop the current going to infinity ?

Click to expand...

I agree with your assessment of that particular scenario.

Similarly, when you connect an ideal Cap to an ideal Inductor 'suddenly', what exists at THAT INSTANT OF TIME to stop the flow of current ?

Click to expand...

I assume you mean an energized ideal capacitor. The back voltage of the inductor expressed by L*dI/dt will reduce the current. The faster the change of current, the more back voltage. If you apply a voltage across a ideal inductor with an ideal voltage source, the current will be a ramp defined by (1/L)*∫v*t over the interval 0 to t. If you connect an inductor to an ideal current source, it will take an infinite voltage from the current source to satisfy its ideality.

Really ? Laplace does that ? I always thought it worked on linear integrals & differentials, not on discontinous functions.

Click to expand...

Yes, the Laplace gives the total response within the interval in which it is defined. The unit step function I used in my analysis was defined for t>0.

So whats the Laplace transform of the Dirac delta then ? I need to know this for a project I'm working on.....

Click to expand...

The Laplace of a Dirac delta is 1.

Secondly - regarding the perfect cap etc etc comment. Is that not what the question was all about ? A perfect cap of 0.4F and a perfect Inductor ( i.e. ZERO resistance - never mind the superconducting point, & zero capacitance) of 1H etc ?? .

Click to expand...

Yes, it was. But only a dead short can give a infinite current with the proviso that no resistance exists.

Ratch