question about parallel RLC circuit.....

newuser

Joined Nov 2, 2004
3
I have a quick question....hopefully someone can help me out here.

If we have a simple parallel LRC circuit with a DC source, what would the voltage be in such a situation?

At DC the Capacitor behaves ideally like an open circuit and the inductor behaves ideally like a short circuit......so what exactly would happen in this situation?

What would the Capacitor/Inductor voltage be?

If anyone can assist me, I would greatly appreciate it.

Thanks.

Maxx

Joined Oct 30, 2004
43
Any paralell DC circuit always assumes a value lower than the lowest resistance in the circuit.

The inductor L in this instance would i guess have an extremely low resistance, say less than a few ohms, so unless the resestor R was lower than this you would have a total resistance value of something close to the inductor.

If it was an AC circuit however, things would be completely different based on the reactance of the components.

newuser

Joined Nov 2, 2004
3
Originally posted by Maxx@Nov 2 2004, 12:56 PM
Any paralell DC circuit always assumes a value lower than the lowest resistance in the circuit.

The inductor L in this instance would i guess have an extremely low resistance, say less than a few ohms, so unless the resestor R was lower than this you would have a total resistance value of something close to the inductor.

If it was an AC circuit however, things would be completely different based on the reactance of the components.
[post=3283]Quoted post[/post]​

Hey Maxx............thanks for responding.

I agree with you about the resistance........so at DC could we say that the inductor effectively shorts out the resistor and capacitor and so there is a very high current through the inductor while the resistor and cap would have a very small current.......

But wouldnt the cap necessarily have (assume the DC source is 12 volts) 12 volts across it after a long time? Since the circuit is parallel...doesnt that imply the inductor would also have 12 volts across it, but with a very large current?

But since v(L) is proportional to d(i(L))/dt then if i(L) is a constant then shouldnt v(L) be zero?

There seems to be a contradiction since we have a capacitor acting like an open circuit at DC in parallel with an inductor acting like a short circuit at DC.

So what would be the voltage across these components at constant DC voltage if they are in parallel?

mozikluv

Joined Jan 22, 2004
1,437
hi,

have you considered cancellation of reactions by the 2 particular components? i might be wrong, am abit rusty about that tank reaction cyberhehe

Joined Oct 6, 2004
61
Originally posted by newuser@Nov 3 2004, 03:28 AM

There seems to be a contradiction since we have a capacitor acting like an open circuit at DC in parallel with an inductor acting like a short circuit at DC.

So what would be the voltage across these components at constant DC voltage if they are in parallel?
[post=3285]Quoted post[/post]​
In any parallel system, Lower one dominates.

Igrecotrei

Joined Sep 15, 2004
7
Originally posted by newuser@Nov 2 2004, 09:28 PM

But since v(L) is proportional to d(i(L))/dt then if i(L) is a constant then shouldnt v(L) be zero?

i(L) is not constant through a coil!!!

arvind

Joined Nov 4, 2004
18
hi,
the statement that cap is dc open and inductor is dc short is correct only at steady state in the dc circuit. before that we have to consider transient . one can analyse using diff equations. do not generalize straight away putting f = 0 for dc. u will get
wrong results for transients. if cap had been dc open for transient as well we would never have seen lc oscillations           Perion

Joined Oct 12, 2004
43
Originally posted by arvind@Nov 4 2004, 11:16 AM
hi,
the statement that cap is dc open and inductor is dc short is correct only at steady state in the dc circuit. before that we have to consider transient . one can analyse using diff equations. do not generalize straight away putting f = 0 for dc. u will get
wrong results for transients. if cap had been dc open for transient as well we would never have seen lc oscillations           [post=3339]Quoted post[/post]​
DC transient responses for RLC circuits gets pretty heavy - second order diff equation in this case if I'm not mistaken.

Perion

newuser

Joined Nov 2, 2004
3
Hey thanks everyone for the thoughts..........

It seems what I originally thought was correct after all..........

In a parallel RLC circuit at DC voltage the inductor will effectively short out and all voltages will be approximately zero with the inductor basically being a short circuit.

In a series RLC circuit at DC voltage the capacitor will effectively act like an open circuit, thus no currect will flow in a series combination.

I over complicated the analysis and began to second guess myself, but it appears my original conclusion was correct.

Thanks

cyberhehe

Joined Oct 6, 2004
61
Originally posted by arvind@Nov 5 2004, 12:16 AM
hi,
the statement that cap is dc open and inductor is dc short is correct only at steady state in the dc circuit. before that we have to consider transient . one can analyse using diff equations. do not generalize straight away putting f = 0 for dc. u will get
wrong results for transients. if cap had been dc open for transient as well we would never have seen lc oscillations           [post=3339]Quoted post[/post]​

We can generalize F=0 at DC in this analysis, since we are talking about DC after all, we dont consider AC impedance here and those imaginary stuffs, not unless you are interested to dig on the fourier details of the transient region. We should take a look at it in a time domain response (and not the frequency response), Caps maybe short and Inductors are open at t=0 , but depending on initial conditions whether you have initial charge stored in the cap or initial energy stored in the inductance which the short open cases might not be true, and let the second order differential equations do the job.

The initial post states an RLC configuration in parallel with a DC source, which did not specify whether this is a DC Voltage or DC Current source. Assuming that it was a DC Voltage source, the analysis may be easier if we convert the Voltage into its equivalent non-ideal current source and combine the shunt source resistance with R to simplify the analysis.

Momentary oscillations maybe there and its response as it approaches steady state may depend on the resistor value in relative to L and C values, depending on the damping case that covers it.

Well anyway, newuser is happy with all those answers as he says, and thats good 