# series L/C ?

Discussion in 'Electronic Basics' started by PDRUNEN, Jan 25, 2005.

1. ### PDRUNENGuest

Hi Group,

I have a series L/C circuit, which will be excited at the resonate frequency
with a 1Vpp Sine Wave.

The reactance of the Inductor is j45 and the Capacitor is -j45 at this
frequency.

The circuit is such that the inductor is connected to the sine source and the
other side to the Cap, the cap is connected here to ground.

If I measure the voltage across the cap at resonate, will it be zero or some
large number?

pdrunen

2. ### Larry BrasfieldGuest

Neither, under the stipulated conditions.

This looks like a homework problem. If you cannot do your
own homework at this early phase of your EE education, I
urge you to find another course of study. It won't get any
easier, and if you have a hard time paying attention to this
simple stuff, the later courses will completely mystify you.

3. ### John PopelishGuest

If the inductor and capacitor were perfect, you could have any finite
voltage across each with essentially zero volts across the pair. Also
if the parts were perfect, it would take infinite current to support 1
volt across them and they would have infinite voltage across each.

Real inductors and capacitors that have AC losses will have finite
voltages across them with a finite voltage across the pair.

4. ### Robert MonsenGuest

The total theoretical impedance is 0. Thus, the only thing limiting the
current is the impedance of the source, and the resistance of the
components.

In order to figure out the voltage between the components, imagine that
there is some resistance R between the inductor and capacitor. Thus,
before capacitor, there is Zl + R, and after it's Zc. You want to figure
out the voltage across the cap, so you use the voltage divider relation:

V = V0 * Zc / (Zc + Zl + R)

Where V0 is the voltage of the driving signal.

Since Zc = -Zl, they cancel, and we have

V = V0 * Zc / R

Zc and V0 are constant at a given frequency.

So, what happens to V as R -> 0?

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

5. ### Andrew HolmeGuest

As John pointed out, real components have losses in the form of series
resistance. Also, real signal generators have source impedance: typically
50 ohms. So, in the real world, your series LC circuit is a series RLC
circuit.

If the source impedance of the signal generator is 50 ohms, and the losses
are 0.1 ohms, the total impedance in the circuit is 50.1 + j45 - j45 = 50.1.
If the source e.m.f. is 1V, the current would be just under 20mA. From
there, you can work out what the voltage drop across each component is.

If the series resistance was much less, the current could be quite large...

6. ### Tom BiasiGuest

The components that your prof put on the blackboard can be considered
perfect components.
The source voltage would have zero internal resistance and the one volt p-p
will be accurate to the infinite decimal place.
The problem was probably designed to force your thinking to phase concepts
or the like; nothing really to solve mathematically.
Tom