# How a series & parallel cap works?

1. Dec 19, 2007

### Pro289

Hi, I have a newbie question regarding caps in series and parallel.

When you apply voltage to a cap in a circuit, is there a difference in the voltage on the other side of the cap between the methods of use above?

Like a cap in series, does it need to be charged fully for the voltage to be seen after the cap? Or will there be a slow rise in voltage? Or is there just instant power after the cap? The same question for a parallel cap. But it seems voltage would just "skip" over a parallel cap and continue on since there's a direct electrical connection to the other end, and there wouldn't be a ramping of voltage.

Could both connection methods be used for "backup" power, so to speak, if the main power is disconnected from the cap?

2. Dec 19, 2007

### capnahab

This reply only in reference to your basic question. To ask this question you already know how resistors work in parallel or series. With caps they work backwards. For example, two 100 ohm resistors in series equal 200 ohms. Two 100 ohm resistors in parallel equal 50 ohms. It is the reverse with caps.

3. Dec 19, 2007

### stewartcs

4. Dec 20, 2007

### Pro289

Those links are fairly informative, but they still didn't answer my question. Let's say I have a circuit with a battery->switch->cap->light. If I turn the switch on, would the light illuminate immediately with full voltage or would you see a slow "ramping" effect as the cap charges? If so, would this be the same for a series cap or a parallel cap?

Also, do both series and parallel caps filter voltage transients on a line the same?

5. Dec 20, 2007

### cepheid

Staff Emeritus
The voltage on the capacitor would "ramp up", approaching the steady state value (the battery voltage) asymptotically (exponentially, in fact). More specifically, if the battery voltage is $V_0$, the capacitance is $C$, and the resistance of the light bulb is $R$ then the voltage across the capacitor will be given by

$$v_C(t) = V_0(1 - e^{-t/RC})$$

If you like differential equations, then you can verity this. Of course, there will only be a current across the capacitor if its voltage is changing (i.e. during the ramp up phase). More specifically, the current through the cap will be given by:

$$i_C(t) = C \frac{dv_C}{dt} = \frac{V_0}{R} e^{-t/RC}$$

So although the voltage across the cap is increasing, it is doing so at an ever decreasing rate, so the current across it decays exponentially. In fact, from this second result, we can see that when the switch is thrown, the light will initially light up at full intensity (as though the cap weren't there), but then it will gradually dim down to nothing.

"Gradually", of course, is misleading. We can see that the current will reach 1/e of its starting value when t = RC. This (RC) is called the time constant, [itex] \tau [/tex]. The circuit probably has a very small time constant. I don't know what the resistance of your typical light bulb is, but let's use values of resistance and capacitance that are common in electronics. Let's say the resistance is on the order of kiloohms, and the capacitance on the order of nanofarads. Then the time constant will be on the order of microseconds, making me wonder if we'd even notice the flash. In fact, I find myself doubting the current will have flown long enough to heat up the filament sufficiently. Can anyone comment on this?

6. Dec 21, 2007

### Pro289

So the light wouldn't "ramp" up, and the cap makes no difference?

7. Dec 21, 2007

### cepheid

Staff Emeritus
Huh? Umm...no? I think I said in quite a bit of detail what it would do in my previous post. I said that it would start off bright immediately and then would dim down until it was completely unlit (once the cap was charged). I also said that this would happen over such a short timescale that it might not happen at all, or even if it did, we may not perceive it. If anything, the circuit might behave as though the lightbulb weren't there, and it was just the cap.