Thank you so much, I will look at KVL again, I get so confused by this because I don't understand how there can be a voltage drop without current flow but of course there must be for the capacitor to have charged. I just remember my professor saying there can't be zero current if there is a voltage drop, is that true?

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

 

Imagine v(t) going smoothly from some negative value to some positive value. Even as v(t) passes through 0 V, it is still changing, and hence i(t) will be non-zero (assuming C is non-zero).

In fact, if the circuit is an AC circuit being driven by a sinusoidal waveform, the current will be a maximum when the voltage is zero and, conversely, will be zero when the voltage is at a max.

I see what you mean, thanks for taking time to explain

 

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

Thank you for the recommended reading, I will check this out!

 

All of the above is good, but I think the OP needs to clarify terminology in his head.

With my students I ask them to use 'potential difference' (PD) when talking about voltages as its correct for any scenario, whereas 'voltage drop' (VD) is, arguably, a dynamic thing relating to the action of, typically, current in a resistor. So VD referring to a battery's output terminals is syntactically incorrect as the battery is the source of the PD. But either VD or PD ACROSS the 35 ohm resistor is clearly and unequivocally as a result of the current through it. To use the term VD when referring to the capacitor is both meaningless and confusing as it plays no part in that. It does, however, experience a PD due to the PD/VD ACROSS the resistor.

In short there can never be a VD across a capacitor, its always a PD reflecting the state of charge of the capacitor. By analogy there is never a physical current flow through a capacitor; there is a change of charge on the capacitor's plates, but electrons never go through the capacitor, so to talk about the current 'through' a capacitor is, IMHO, technically meaningless.

I like this, I need to try and think more about the potential difference, your explanation makes a lot of sense, thank you

 

Even if I were to stipulate to everything you've just claimed, the OP also needs to be able to work in the real world using terms as they are used in the real world. In the real world, "voltage drop" and "potential difference" are almost always used synonymously.

As for the current through a capacitor, yes, electrons don't flow through the capacitor's dielectric, but the displacement current does.

That's a good point, I've not come across displacement current on my course yet

 

That's a good point, I've not come across displacement current on my course yet

And you won't any time soon.

[simple explanation before someone jumps down my throat!]

Its a very subtle extension of Maxwells electromagnetic equations to take account of the fact that a rapidly changing electric field between the plates of a capacitor must give rise to a changing magnetic field around that space, in the same way that the changing current in the capacitor leads gives rise to a magnetic field around them (else the magnetic field would collapse at the capacitor plates and lead to all sorts of inequalities). The displacement current is the theoretical current that would need to flow in that space to generate the magnetic field. It is vanishingly small and has little practical value in any common circuit, but could be an issue at very high frequencies and voltages where its potentially a source of noise due to magnetic induction in surrounding components.

 

And you won't any time soon.

[simple explanation before someone jumps down my throat!]

Its a very subtle extension of Maxwells electromagnetic equations to take account of the fact that a rapidly changing electric field between the plates of a capacitor must give rise to a changing magnetic field around that space, in the same way that the changing current in the capacitor leads gives rise to a magnetic field around them (else the magnetic field would collapse at the capacitor plates and lead to all sorts of inequalities). The displacement current is the theoretical current that would need to flow in that space to generate the magnetic field. It is vanishingly small and has little practical value in any common circuit, but could be an issue at very high frequencies and voltages where its potentially a source of noise due to magnetic induction in surrounding components.

Thanks for that explanation, one day I will hopefully understand the intricacies of how this works! Everything I learn about electronics and electrical engineering seems to be the tip of another iceberg, its fascinating but scary at the same time

 

That's a good point, I've not come across displacement current on my course yet

Depending on what courses you take, you may not ever come across it.

It is more properly known as Maxwell's displacement current and is actually Maxwell's contribution to Maxwell's equations -- but it is considered one of the most important contributions and a turning point in our understanding of electromagnetic phenomena.

It is, in simple terms, a recognition that a changing electric field, like an electric current, is a source of a magnetic field.

 

Thanks for that explanation, one day I will hopefully understand the intricacies of how this works! Everything I learn about electronics and electrical engineering seems to be the tip of another iceberg, its fascinating but scary at the same time

I've been doing it for 55 years across many fields and many industries and I'm still learning new things. Fortunately I've got the basics sorted lol

 

And you won't any time soon.

[simple explanation before someone jumps down my throat!]

Its a very subtle extension of Maxwells electromagnetic equations to take account of the fact that a rapidly changing electric field between the plates of a capacitor must give rise to a changing magnetic field around that space, in the same way that the changing current in the capacitor leads gives rise to a magnetic field around them (else the magnetic field would collapse at the capacitor plates and lead to all sorts of inequalities). The displacement current is the theoretical current that would need to flow in that space to generate the magnetic field. It is vanishingly small and has little practical value in any common circuit, but could be an issue at very high frequencies and voltages where its potentially a source of noise due to magnetic induction in surrounding components.

If the displacement current in a capacitor's dielectric is "vanishingly small", then that requires that the conduction current in one side and out the other are also vanishingly small since they are identical.

 

Depending on what courses you take, you may not ever come across it.

It is more properly known as Maxwell's displacement current and is actually Maxwell's contribution to Maxwell's equations -- but it is considered one of the most important contributions and a turning point in our understanding of electromagnetic phenomena.

It is, in simple terms, a recognition that a changing electric field, like an electric current, is a source of a magnetic field.

I just reviewed our course catalogue and it looks like semester 2 of 3rd year covers this, I don't know what depth they go into but it looks complex!

 

I've been doing it for 55 years across many fields and many industries and I'm still learning new things. Fortunately I've got the basics sorted lol

That's incredible, hopefully I'll get there someday too!

 

That's incredible, hopefully I'll get there someday too!

My one bit of advice - all the theory in the world is useless without practical experience; real learning comes from doing. I built my first 'radio' aged 7, various electronics projects through my teens and the rest has been both a career and a hobby ever since. IMHO you can only get an appreciation of how stuff works by doing it for real, though I'll admit that circuit simulators like LTSpice now make it much easier to 'have a go' without getting physical though learning to build projects, soldering, fault-finding, etc are useful skills too.

 

hi Mojo,
Look at this redraw, does it make it clear.
E

Hi Eric,

That redraw does not look equivalent.
Open circuit the cap, erase it, and see what results: 35 and 5 in series, and that set in parallel with 20.
The voltage across that set can then be calculated, then use the voltage divider formula to get the voltage at the voltage across the cap.

 

h i Al,
Read post #13, my lucky number.
E