# Understanding capacitor behaviour (fully !)

Dear forum

I am a school teacher, and whilst I KNOW the behaviour of a capacitor, I must admit I do not fully UNDERSTAND its behaviour.

Please see the left hand image – a capacitor and resistor in parallel over a battery:

http://www.mpklein.co.uk/cap_query.jpg

Textbooks will tell you that, initially, the voltage over the capacitor is zero (due to it having zero charge). Because resistor B is in parallel , the voltage across resistor B must also be zero. Therefore no current can flow through resistor B.

One problem I have with this is why the initial voltage across the capacitor must be zero. Surely if the spacing between the plates was increased a little (and perhaps the plate area reduced) this is equivalent to the other circuit – essentially a gap. Indeed, wouldn't the charge stored on a pair of leads, with a gap, be zero? Wouldn’t a gap in the circuit have the same voltage across it as the battery? How does the capacitor achieve zero volts despite being wired across the battery?

I also do not understand this logic: the voltage across the capacitor is zero, so because they are in parallel the voltage across resistor B is zero . Why is this? The capacitor is in parallel with the battery, so shouldn’t it have the same voltage across it as the battery?

The way I see it we have one route with a gap (the capacitor) and another route which should conduct (the resistor) – so why don’t we get a current through the resistor ? Please help !

Many thanks

Matt

dlgoff
Gold Member
Once the capacitor is charged, it will have the same voltage across it as the resistor does. Here's a useful link on http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html" [Broken].

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Delta2
Homework Helper
Gold Member
A gap is like a capacitor which has very small capacitance C. So it is charged too fast (practically zero time, charging time is analogous to RC) to the desired voltage and the charge accumulated Q is too small (practically zero). Voltage as a ratio of Q/C=(0/0) can be finite though. So for the gap u assume that instantenously reaches its final voltage V.

For a capacitor with not so small capacitance it ll take some time to be charged to the voltage V. So it starts from zero voltage and keep charging to the final V. In the examples u give the final V is $$V=R_B\frac{E}{R_A+R_B}$$ for both, it just takes more time and more charge accumulated in the left example.

There is current flowing through resistor B but at t=0 is zero. As charges build up on capacitor they create a voltage so current will flow through resistor B. The final voltage when capacitor is fully charged would be V as stated above and the final current will be $$\frac{V}{R_B}$$.

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Delta2
Homework Helper
Gold Member
Once the capacitor is charged, it will have the same voltage across it as the resistor does. Here's a useful link on http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html" [Broken].

Capacitor will always have the same voltage as resistor B, you mean that once it is charged there would be no current flowing through it and no further charge accumulated on its plates.

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Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0. The answer is because the effective resistance of the capacitor @ t=0 is zero.

Consider two resistors, R1 & R2 in parallel. If R1 = 0 ohms and R2 = 100 ohms, then the voltage across both resistors is E = I * (1/(1/R1 + 1/R2)) ==> So for R1 = 0, E = 0 and I is infinite. This is similar to a capacitor at t=0.

Obviously a capacitor is NOT a variable resistor, but it may help to think of it that way.

Fish

rcgldr
Homework Helper
Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0.
It might be simpler to explain that prior to t=0 (the starting time), the battery is not connected to the circuit, and that at t=0 the battery is initially connected to the circuit, when the capacitor voltage is still initially zero.

Delta2
Homework Helper
Gold Member
Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0. The answer is because the effective resistance of the capacitor @ t=0 is zero.

Fish

The voltage is zero because the capacitor will have zero charge at t=0 hence its voltage V=Q/C=0/C would be zero. This means that there are no charges in its plates to oppose the incoming of further charges, hence one could say its effective resistance is 0.

The voltage is zero because the capacitor will have zero charge at t=0 hence its voltage V=Q/C=0/C would be zero. This means that there are no charges in its plates to oppose the incoming of further charges, hence one could say its effective resistance is 0.

Thank you for your reply Delta-squared. But if we just had a gap in the circuit, would this not store zero charge on the ends of the wires and so have zero voltage across?

Actually, I think what he is asking is why is there zero voltage across the resistor @ t=0. The answer is because the effective resistance of the capacitor @ t=0 is zero.

Consider two resistors, R1 & R2 in parallel. If R1 = 0 ohms and R2 = 100 ohms, then the voltage across both resistors is E = I * (1/(1/R1 + 1/R2)) ==> So for R1 = 0, E = 0 and I is infinite. This is similar to a capacitor at t=0.

Obviously a capacitor is NOT a variable resistor, but it may help to think of it that way.

Fish

Thank you for the Fish 4 Fun. Whereas a GAP in the circuit would have effectively an infinite resistance?

A gap is like a capacitor which has very small capacitance C. So it is charged too fast (practically zero time, charging time is analogous to RC) to the desired voltage and the charge accumulated Q is too small (practically zero).

This is very useful, thanks.

Whereas a GAP in the circuit would have effectively an infinite resistance?

Exactly. I thought that explanation might bring it "home" for you. It is technically wrong, but functionally accurate.

In understanding reactive components and impedance, it helps to think of V and I rather than focusing on "resistance". For instance, in the case of a capacitor:

I = E/R * e^(-t/RC)

Where "R" is some value greater than 0, even if it is very small. As R approaches 0, the current approaches infinity. What this implies is that there is always some "R", even if it is in the power supply itself (this is referred to as "source impedance") that prevents the current from actually being infinite. To get a "feel" for this, you might want to play with very small values of R here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

Capacitors in a DC circuit are generally only "interesting" for finite periods of time after the power is switched on or off.

Hope this helps,

Fish

Thank you for your reply Delta-squared. But if we just had a gap in the circuit, would this not store zero charge on the ends of the wires and so have zero voltage across?

That was answered in posts #2 and #3. A gap stores nearly zero charge so that the voltage is built up very quickly. A gap is also almost immediately discharged.

If you ever did some soldering on audio circuits, you may know that a big capacitor is needed to reduce hum that is due to power supply voltage variations. Inversely, a gap (no capacitor) results in a big hum. If you look at the link of posts #2 and #11 you may understand why. OK, well I think I am there now. Capacitors have always puzzled me a bit, but less now !

Thank you for everybody's comments, and the links too. You have all been a great help.

Matt

Delta2
Homework Helper
Gold Member
Thank you for your reply Delta-squared. But if we just had a gap in the circuit, would this not store zero charge on the ends of the wires and so have zero voltage across?

gap stores almost zero charge Q but has almost zero capacitance C, so the ratio Q/C which is the voltage can be finite and non zero. For example if Q=0.0001 and C=0.00001 then Q/C=10

dlgoff
Gold Member
Capacitor will always have the same voltage as resistor B, you mean that once it is charged there would be no current flowing through it and no further charge accumulated on its plates.
Yes.

rcgldr
Homework Helper
On a side note, is it possible to plot a current versus time graph for a fixed voltage supply + capacitor without a resistor, and caculate the charge versus time? This would only be possible if the area under this curve from t = zero to t = some finite time isn't infinite, despite the fact that the initial current is infinite for an instant in time.

Delta2
Homework Helper
Gold Member
On a side note, is it possible to plot a current versus time graph for a fixed voltage supply + capacitor without a resistor, and caculate the charge versus time? This would only be possible if the area under this curve from t = zero to t = some finite time isn't infinite, despite the fact that the initial current is infinite for an instant in time.

hmmm, in this case current $$I=VC\delta(t)$$ and

$$q=\int_{0}^{\infty}VC\delta(t)dt=VC$$

but we just cant have instanteneous charge of the capacitor there would be other factors limiting the speed of charging (like the distance of capacitor from the source, the e-field which makes free electron to move doesnt travel instanteneously) which are not taken into account by circuit theory.

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The capacitor is in parallel with the battery, so shouldn’t it have the same voltage across it as the battery?

What's in the box labeled "A"? The capacitor is NOT in parallel with the battery, because of the existence of device "A".

Immediately after charging begins, the capacitor looks like a short circuit, because charges flow onto one capacitor plate and flow off of the other capacitor plate so rapidly that there appears to be a shorted path "through" the capacitor, and for an instant no current at all flows through "B". As a result, the entire battery voltage drop is across "A".

What's in the box labeled "A"? The capacitor is NOT in parallel with the battery, because of the existence of device "A".

Immediately after charging begins, the capacitor looks like a short circuit, because charges flow onto one capacitor plate and flow off of the other capacitor plate so rapidly that there appears to be a shorted path "through" the capacitor, and for an instant no current at all flows through "B". As a result, the entire battery voltage drop is across "A".

Hi Mikelepore

A & B are resistors.

You are right, the capacitor is not directly parallel to the cell , I forgot about that.

What would happen if resistor A was not included in the circuit? Even if the capacitor was wired directly over the battery it would initially still have zero volts across its terminals wouldn't it ? This is unusual, because most other components would have the battery voltage across them.

What would happen if resistor A was not included in the circuit? Even if the capacitor was wired directly over the battery it would initially still have zero volts across its terminals wouldn't it ? This is unusual, because most other components would have the battery voltage across them.

The instant an ideal capacitor is connected to ideal battery terminals it will have battery voltage across it. Of course in the real world there is still a tiny resistance represented by A, such as the resistance in the capacitor plates, the battery and the connecting wires, so the jump is not a perfect step function. In other words really fast, but slower than the answer above from Delta with just wire ends.

q = it = CV are convenient relationships to remember...

that is, the charge (q) on a capacitor equals the current (i) that has passed thru the capacitor times the duration of current flow (t) , which equals the capacitances (C) time the voltage (V) across the capacitor.

With a capacitor, current starts off very high, then slows to zero as charge (electrons) build up on the capacitor plate and new electrons have nowhere to go...new electrons are opposed by those already on the capacitor plate.

Suppose you place two capacitors of different size (different capacitance, C) in parallel across the same battery terminals...from the above q=CV you can see this means the two will have different q because C is different and V must be the same....then from the other q = it, you can tell infer either i was different or one took longer to charge or both these occurred.

What would happen if resistor A was not included in the circuit? Even if the capacitor was wired directly over the battery it would initially still have zero volts across its terminals wouldn't it ?[

If resistor A were not included, then whatever small resistance is in the wiring, even if it's a small fraction of one ohm, would be the R in the expression for the time constant, the product RC. That's why the capacitor would appear to fully charge almost instantly, because of that smallness of that RC product. However, for an instant after closing the switch, the battery voltage would be fully across that wiring resistance, and none of it across the capacitor.

In an ideal mathematical model, if you omit the series resistor and also let the wiring resistance be zero, you can safely discuss the steady state condition when no switch has been thrown in a long time, but a discussion of the moment of switching "blows up", because for an instant the capacitor acts like a short, the zero resistance would conduct current V/R and dissipate power V^2/R, and there are zeroes in the denominators.

This is unusual, because most other components would have the battery voltage across them.

I don't undertand the last sentence.

I'm also reviewing this subject. I'm teaching this too, and I'm going to deliver this lesson in a couple weeks from now.