Can I Charge A Capacitor While Using It?

1. Jul 15, 2010

ijrdn

The thing between A and B is an inductor, don't know how to draw the coil :tongue:
Im wondering if theres AC currents flowing through point A and B?
Or is the capacitor keep charging up w/o discharging?
Or What happens???????

If that doesn't charge the capacitor in the oscillator, I want to know how to.

Thanks

Last edited: Jul 15, 2010
2. Jul 15, 2010

gnurf

Charging a capacitor is using it. It charges and discharges, storing/releasing energy in/from the electric field between its plates.

Where would the AC current in your circuit arise from? You have a battery (which is a DC source) which is shorted with your wire, effectively bypassing your capacitor, which you might as well remove from the circuit---it does nothing as your circuit stands. If you didn't short it out, your capacitor would charge up with its characteristic exponential curve leveling off at the battery voltage, at which point no more current would flow in the circuit.

3. Jul 15, 2010

ijrdn

Sorry, I was being a total idiot, I forgot to draw the inductor.

4. Jul 16, 2010

gnurf

Look up the expression for the impedance of an inductor (L), and you'll see that it acts as a short to DC. So no, there's no AC current between A and B.

To charge the capacitor (C), remove the L (and thus the DC short-circuit), and wait until the C charges up to the battery voltage. At this point you can remove the battery and your C will contain a certain amount of energy in the form of an electric field between its plates.

If you were to connect a L over the C at this stage (when the battery is removed), the stored energy would ideally slosh back and forth between C and L (which stores energy in a magnetic field) indefinitely.

5. Jul 16, 2010

gnurf

Also, the circuit symbol for an inductor looks like this:

[PLAIN]http://www.kpsec.freeuk.com/symbols/inductor.gif [Broken]

Currently, your attempt at drawing a inductor looks more like a resistor to me. Though, I guess that's better than a straight line as you drew the first time :)

Last edited by a moderator: May 4, 2017
6. Jul 16, 2010

ijrdn

Lol, I tried to draw that, but I failed XD, so I just stuck to the "resistor" thingy.
Just one more question, how do you charge the capacitor in an oscillator circuit??

7. Jul 16, 2010

gnurf

That's like saying "I tried to say thank you, but I couldn't find the words, so I said f*ck you instead". Symbols, like words, have meaning ... (most of the time anyway).

On charging the capacitor: Once the battery is removed the capacitor's potential difference, V=q/C, will initially drive the charge off the capacitor, which means a current will begin to flow in the LC circuit. The flow of current is key, because once the capacitor is completely discharged, how does the oscillator keep going?

The answer lies in how a inductor reacts to the (rate of) change in current passing through it. As you should know, the rate of change in current is proportional to the voltage across the inductor, V = L di/dt, which means that the current can't just suddenly (as in di/dt --> inf) stop, as that would require an infinite voltage across the inductor. Instead, the inductor induces an emf that is in the opposite direction of the current, effectively reversing the polarity of the capacitor which then recharges... and so on. Got it?

8. Jul 16, 2010

ijrdn

Yea I got that, guess I don't know how to ask a question......
How do you connect a battery to an oscillator circuit to give it that extra "juice" keep it going instead of dying by itself after it loses all of its energy due to resistance.

9. Jul 16, 2010

gnurf

There are several ways to do it, I guess. E.g., the circuit below is a Colpitts Oscillator:

The oscillation frequency is:

[PLAIN]http://images.wikia.com/wikitex/images/b/bc/bc8/93f498a73f8ed418ee94a02e814fd1.png [Broken]

You'd need to bias the transistor and add a power source, but luckily they've added the spice diagram for you to play with: