# Capacitance and frequency

1. Mar 28, 2004

### chamsy

Hi,

I have learned that as a capacitor discharges it emits some kind of a wave. Plz tell me wether the frequency of this wave increases as the capcitance of this capacitor increases. Or plz tell me is there a way to find out the cpacitance of a capacitor by looking at the wave that this capacitor emits.

Thanks a lot,
Chamal.

2. Mar 28, 2004

### Chen

I learned that when a capacitor discharges it causes a current to flow through it. I also learned that in AC circuits its capacitive reactance is inversely proportional to the frequency of the source, and that's about everything I know about the connection between capacitance and frequency.

Last edited: Mar 28, 2004
3. Mar 28, 2004

### Integral

Staff Emeritus
My best guess would be no. The frequency of any emitted EM will be determined by the frequency at which the circuit is running. If you change the operation frequency the circuit capacitance will remain constant but emitted frequencies will change.

4. Feb 4, 2010

### sumair quresh

I am actually looking for an explanation not mathematical to understand the logic behind the decrease in the capacitance with the increase in frequency.
Can any one help me??

BEST REGARDS
SUMAIR QURESHI

5. Feb 4, 2010

### torquil

Actually, the capacitance is constant (ideally). It is the resulting impedance that varies with frequency. Just check it out e.g. on wikipedia.

Torquil

6. Feb 4, 2010

### Bob S

The resonant frequency of an electrical circuit with a capacitance C and an inductance L is

f= (1/2 pi) sqrt(1/LC) Hertz

So when the capacitance deceases, the frequency increases.

Bob S

7. Feb 4, 2010

### sophiecentaur

That's a bit like asking for your bank statement in non-numerical terms. The sums tend to go with the territory.
The basic equations are not hard and give a far better idea of what's going on than arm waving. I suppose the simplest way of explaining how the (single) electromagnetic pulse is generated from a capacitor discharging through a resistor is that a small capacitor will have less charge in it, for a given voltage so it will take less time to discharge. Consequently, the resulting pulse will be shorter. The current flowing around the circuit and the changing voltage will generate the field associated with the em pulse.
If an inductor is included in the circuit, the waveform will be a damped oscillation rather than a single pulse. The oscillation is due to the energy moving from the charged capacitor to the magnetic energy around the inductor and back again - ad infinitum - except that it will decay due to resistance and energy being radiated away. The lower the values of Capacitor and inductor, the quicker the oscillation will be - by the same sort of argument as above. But the Maths says it a lot more elegantly - why not bite the bullet?

8. Feb 4, 2010

### sophiecentaur

As for the other question - it is quite possible to measure the value of a capacitor by seeing how it affects the resonant frequency of a tuned circuit. The old fashioned Q meter did just this in order to measure component values. Cheap and cheerful and very easy to use.
Deja vu - I just included this fact in another post the other day.

9. Feb 5, 2010

### Naty1

Perhaps what you mean is that when a charged capacitor discharges, and it does so with some resistance in the circuit, say the resistance of the wire, the discharge is not instantaneous....it's governed by the time constant of the circuit.

If there is no resistance, in an ideal circuit the discharge is instantaneous...a step function...

If there is a tiny inductance assumed, then there is a brief resonant time period during the capacitor discharges as outlined by BobS. It's a rapidly decreasing sinusoidal wave rather than a smooth curve with just R and C.

Yet another way to look at your question is to note that a varying voltage produces and electromagnetic wave...this would be studied via an entirely different set of equations, those of Maxwell....for fields in a vacuum rather than in the circuit wiring and electrical components as measured with a voltmeter or ammeter.

10. Feb 5, 2010

### Pythagorean

Depending on how your capacitor relates to the resonant frequency of the circuit, the discharge may cause a distortion in the shape of the waveform. In this case, it's actually a band of frequencies, rather than a single frequency.

You could find this band by doing a Fourier transform on the signal.