# [beginner] Current in a capacitive circuit

Hey people!.. this is what i've been wondering for a while and couldn't come up with anything (asked my frnds too)

It has been derived in my textbook that when AC is applied to a pure capacitive circuit (i.e. contains only a capacitor)... current leads ahead of the emf by phase difference of pi/2.
Does that imply that even before i "switch on", current will already be flowing? that does sound absurd to me... how can current flow without applying emf....

a better explanation on lagging and leading of phase might help too..
Thank you

PS: I mistakenly typed the topic name as current in "inductive" circuit... sorry!... moderators plz help me change it if its possible.....

Last edited:

I note you state beginner.

So please tell us if your studies have covered what happens to a capacitor when you apply DC to it, ie have you seen the (exponential) curves for the rate of growth of voltage across a capacitor?

If you have seen these it will make the explanation much easier.

Last edited by a moderator:
no i havent seen those curves
i have a rough idea of what happens when capacitor is connected to DC circuit... temme if m correct..
When DC is applied to capacitive circuit, current flows for a very short time till the capacitor is filled and then there is no current in the circuit as the capacitor and cell become equipotential....

Last edited by a moderator:
berkeman
Mentor
Hey people!.. this is what i've been wondering for a while and couldn't come up with anything (asked my frnds too)

It has been derived in my textbook that when AC is applied to a pure capacitive circuit (i.e. contains only a capacitor)... current leads ahead of the emf by phase difference of pi/2.
Does that imply that even before i "switch on", current will already be flowing? that does sound absurd to me... how can current flow without applying emf....

a better explanation on lagging and leading of phase might help too..
Thank you

PS: I mistakenly typed the topic name as current in "inductive" circuit... sorry!... moderators plz help me change it if its possible.....

(I fixed the title)

Do you know the differential equation that relates the voltage across a capacitor to the current through it? That's really the key to understanding the phase lead/lag stuff.

no i havent seen those curves
i have a rough idea of what happens when capacitor is connected to DC circuit... temme if m correct..
When DC is applied to capacitive circuit, current flows for a very short time till the capacitor is filled and then there is no current in the circuit as the capacitor and cell become equipotential....

Exactly so.

When you first switch on there is zero voltage across the capacitor, but a large current flows.

As time proceeds the voltage across the capacitor increases and the current decreases.

This is what is meant by the current leads the voltage. Exactly the same thing happens within an AC cycle, although it obviously repeats itself over and over again at the AC frequency.

If I have time later I can post some curves, or perhaps someone else will do it.

Does this help?

go well

@studiot
totally helped.. thanks a ton... but i'd like to see those curves.... wanna see if they are like how i imagine them to be....

Could you also explain with that simplicity why current would lag behind the emf in an inductive circuit?... that'll help a lot sorting out my confusion about in which circuit it lags n in which it leads and i wont have to mug it up....

also, often i see when i switch off my speakers or monitor, the monitor/speaker light stays on for quite a while (i am talking about switching off from the mains not the power button) has it got to do something with current leading d emf? if so, what n why?

@berkeman
thank you for fixing the title....
no, i am not aware of the differential equation but i'd really love to know them... i'd be grateful if you could post them with a lil explanation..

I thank both of you for replying so quickly and helping me out!

berkeman
Mentor
@studiot
totally helped.. thanks a ton... but i'd like to see those curves.... wanna see if they are like how i imagine them to be....

Could you also explain with that simplicity why current would lag behind the emf in an inductive circuit?... that'll help a lot sorting out my confusion about in which circuit it lags n in which it leads and i wont have to mug it up....

also, often i see when i switch off my speakers or monitor, the monitor/speaker light stays on for quite a while (i am talking about switching off from the mains not the power button) has it got to do something with current leading d emf? if so, what n why?

@berkeman
thank you for fixing the title....
no, i am not aware of the differential equation but i'd really love to know them... i'd be grateful if you could post them with a lil explanation..

I thank both of you for replying so quickly and helping me out!

The equation for a capacitor is:

$$I(t) = C \frac{dV(t)}{dt}$$

The equation for an inductor is:

$$V(t) = L \frac{dI(t)}{dt}$$

If you aren't familiar yet with differential notation, the dV/dt just stands for the change (or delta) in voltage with respect to time. When you differentiate V(t) with respect to time, you get the derivative dV(t)/dt.

So from the equation for the capacitor, you can see how centaur's description of the relationship between current and voltage comes about. If you all of a sudden try to change the voltage across a capacitor quickly, that takes a large current. And if you have a current flowing through the capacitor, that means that the cap voltage has to be changing.

Also note how the position of the voltage and current are interchanged in the two equations for the capacitor and inductor. That results in the opposite relationship for lead/lag of voltage and current for caps versus inductors.

@berkeman
yeah that did help a lot

i was familiar with the differential equation for inductance.. just didnt know that it is called that

however could you temme how do you get that differential equation for capacitance? please!

berkeman
Mentor
@berkeman
yeah that did help a lot

i was familiar with the differential equation for inductance.. just didnt know that it is called that

however could you temme how do you get that differential equation for capacitance? please!

I believe it comes from the definition of capacitance:

$$C = Q/V$$

The more charge Q that is stored on the capacitance C for a given voltage, the larger the capacitance is.

So re-arrange the equation and differentiate both sides:

$$Q = CV$$

$$\frac{dQ}{dt} = C \frac{dV}{dt}$$

(capacitance C is constant, so comes out before the differentiation)

And since the current is defined as the flow of charge:

$$I(t) = \frac{dQ}{dt} = C \frac{dV(t)}{dt}$$

.

thanks!..
now just to summarize... tell me if I'm right..
In a capacitive circuit, for a small change in voltage, there is a large change in current... so current is always more in the circuit... thus current is always leading the voltage

for a inductor, a small change in current results in large change in voltage... so voltage is always more in the circuit... thus voltage is leading the current implying current lags behind the voltage

that's it right?

Last edited by a moderator:
berkeman
Mentor
thanks!..
now just to summarize... tell me if I'm right..
In a capacitive circuit, for a small change in voltage, there is a large change in current... so current is always more in the circuit... thus current is always leading the voltage

for a inductor, a small change in current results in large change in voltage... so voltage is always more in the circuit... thus voltage is leading the current implying current lags behind the voltage

that's it right?

No, not right.

In a capacitor, the current is proportional to the changes in the voltage.

In an inductor, the voltage is proportional to the changes in the current.

.

yeah i get that they are proportional.... what i dont understand is how does current proportional to change in voltage and voltage proportional to change in current suggest whether current lags or leads.... kindly elaborate.........

i kinda understood what studiot said about capacitors... can a similar explanation be given in case of inductive circuits?

Last edited by a moderator:
OK, as promised, I have posted a sketch of the graphs for charging a capacitor, C.
I have shown the voltage and current curves above each other on the same timescale for comparison.
I have also included a resistor, R. If we do not do this the current starts off at theoretical infinity, certainly with a large spark or worse.
This is what happens if we close the switch at time t=0.

You can see the shapes of the curves and I have included their equations. E is the battery voltage.

These equations are the solutions to the differential equations Berkeman posted.

You should check this for yourself by differentiating them and substituting.

You can see how the current starts high and decreases to an asymptote at t = infinity of zero, whilst the voltage starts at zero and increases, ever more slowly, to an asymptote equal to the battery voltage.

As regards to changing the capacitance for an inductance, how about you have a go first off?

#### Attachments

• capDC.jpg
8.6 KB · Views: 800
berkeman
Mentor
yeah i get that they are proportional.... what i dont understand is how does current proportional to change in voltage and voltage proportional to change in current suggest whether current lags or leads.... kindly elaborate.........

i kinda understood what studiot said about capacitors... can a similar explanation be given in case of inductive circuits?

For capacitors, voltage lags the current in time.

Let the voltage V(t) waveform be a sine wave, then the current is proportional to dV(t)/dt, which will be a cosine wave. You can see from the plots above that the sine wave is 90 degrees behind the cosine wave in time...

hey studiot... thanks a lot... the graphs helped
hey berkeman... thanks a lot to you too..... there is no confusion now

about the capacitor, i have a clear idea now... thanks to you both!

FOR INDUCTIVE CIRCUIT:
I must always be a sine wave... therefore e will be cosine wave... thus i lags behind... got that...

but what would ACTUALLY be happening in an inductive circuit that makes the current lag
is it because of the back emf that the inductor produces? if so, please elaborate on that

also when you say emf lags behind... which emf lags behind? the applied emf(cell) or the back emf due to self inductance?

Berkeman has shown that the

capacitor current = the slope of the voltage time graph (x a constant)

Do you understand this? Can you see this on my graphs?

So when the slope is high at the beginning the current is high and as the slope decreases the current decreases, until when the slope is zero the current is zero.

This is why you have (probably) been taught calculus in maths classes.

Berkeman has also shown that

inductor voltage = the slope of the current time graph (x a constant)

It would solidify your understanding if you could draw or guess at similar graphs for an inductor.

Have a try - it doesn't matter if you get it wrong -we will help.

As a starter, if you replace the capacitor with an inductor in my circuit, when you first close the switch you are suddenly applying the full battery voltage, E. So dV/dt is large.

in an inductor... with time current will increase and voltage will decrease?
so the current-time graph of the capacitor will look like the voltage-time graph of the inductor?

Good try.

The current increases, rapidly at first then ever more slowly towards some limiting value, like the voltage time graph for the capacitor.

The voltage is more tricky. Before I answer in detail I meant to ask have you heard of back emf? Have you covered inductor action yet in your studies?

but what would ACTUALLY be happening in an inductive circuit that makes the current lag
is it because of the back emf that the inductor produces? if so, please elaborate on that

also when you say emf lags behind... which emf lags behind? the applied emf(cell) or the back emf due to self inductance?

yeah i know about the inductor a little... it stores electrical energy in terms of magnetic field... yes i know what back emf is... it is the emf that opposes the applied emf due to self inductance?
i had also mentioned about whether back emf is the cause of current lagging before.....

tell me if i am right
through an inductor when AC current flows... because of ampere's law magnetic field is created... but since the current is always changing, the magnetic field is always changing... now due to faraday's law... the changing magnetic field causes a emf.... but according the lenz law the emf must oppose the cause of it which is the applied alternating emf.... this emf which opposes the applied emf is called back emf?

please tell me if its right... i really want to get my concepts cleared

Last edited by a moderator: