# Ohm's law

When plotting potential difference VS current (volts VS amps).....how does one explain the intercept at (0,0) ?

Is it because without potential difference you can't have current and vice versa? Is that true all the time?

Thanks

Related Introductory Physics Homework Help News on Phys.org
chroot
Staff Emeritus
Gold Member
Potential difference provides the "force" which drives current. Without a difference in potential, there is no movement of charge (i.e. current).

- Warren

Warren can you explain why sometimes the current can lead voltage in (I think) AC circuits? Its atleast kind of related to the thread..

chroot
Staff Emeritus
Gold Member
You mean, why aren't the voltage and current waveforms in phase for reactive circuits (those with capacitors or inductors)?

- Warren

Thats probably what I'm talking about . I never understood how hte current can flow before the potential difference is in place.

Integral
Staff Emeritus
Gold Member
ELI the ICE man

E= Voltage,
L= Inductance
I = Current
C= Capacitance

ELI means L for inductance E comes first so Voltage leads current in an inductor.

ICE, C for capacitance , Current comes first so Current leads voltage in a Cap.

To understand why this is you need to consider the physical reality of how the devices work. Initially a cap is a low resistance to current as the plates charge the current drops off. Thus the initial current is high.

In an inductor the initial must build the electric field, thus must do work and is restricted early. So the initial current is low.

I am sure Warren will want to improve on this cliff note version but this is a starting point.

Integral
Staff Emeritus
Gold Member
Whozum,
The p.d. is in place and a circuit must exist.

This is referring to the current and voltage drop across the DEVICE. So initially the current through a cap is high, but the voltage drop is low. Once the cap is charged the current is low but the voltage drop is high.

For an inductor, while the field is being "built" the current through the inductor is low but the voltage drop across the device is high.

Ok, I understand the capacitor completely but I'm going to repeat what you said just to make sure i can explain it.

A simple circuit with just a capacitor and a battery with a potential difference of 12V between its plates. An electron leaves the terminal due to the PD of 12V, here, the PD is causing the current flow, so we have ELI. once it gets to the capacitor the current reaches the first plate with still 12V potential difference to go through before reaching the battery's positive terminal. (The capacitor has 0 PD between its plates before the electron gets there, correct?)

Inside the capacitor, there is a build up of charge on the terminal and this causes a potnetial difference between it and its counterplate, giving us the ICE, since the presence of current increases the PD. As this process continues, the current goes to 0 and the PD between the capacitor's plates increases to 12V asymptotically.

I'm going to review inductance now since I really never learned it in the first place. Can someone please check the accuracy of the above?

chroot
Staff Emeritus
Gold Member
whozum,

There's a lot wrong there, honestly. To start with, when you connect a battery to a capacitor, the electric field propagates through the wires at nearly the speed of light. It has nothing to do with the movement of any electrons; electrons, in fact, drift only very slowly through a circuit. The electric field which propels the electrons, however, propagates at near light speed.

The current is not like a gush of water coming down a mountain, moving from one place to another -- the current is like a conveyor belt that gets "turned on" everywhere at once in the circuit as the emf (voltage, potential difference) is established through it.

Let's not talk about batteries so much, since batteries are DC sources. The difference in the pahse of emf and current only manifests itself in an AC system, where both are constantly changing in sinusoidal fashion.

First, make sure you understand one fact about a sinusoidal voltage source: the voltage changes most rapidly through the zero-crossing. (Look at the zero-crossings of a sine wave if you don't immediately recognize this fact.)

You also hopefully know that the current into a capacitor is proportional to the rate of change of the voltage across it. In other words, the current into the capacitor is largest right when the voltage is zero. The current into the capacitor is largest at the zero-crossings of the source.

In other words, when the voltage across a capacitor in an AC system is zero, the current into that capacitor is at a maximum. The voltage and current waveforms are therefore not in phase.

To understand this a bit better mathematically, consider the relationship I already mentioned:

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

Now, if V is a sine wave, I depends on the derivative of that sine wave. And... what's the derivative of a sine wave? A cosine wave -- same amplitude, 90 degrees different in phase. Current leads voltage by 90 degrees in a capacitor.

The same argument can be used for an inductor, which is described the relationship:

$$V = L \frac{dI}{dt}$$

Here, if V is a sine wave, I must be a negative cosine wave -- they are 270 degrees different. In an inductor, the voltage leads the current by 90 degrees.

- Warren

The current is not like a gush of water coming down a mountain, moving from one place to another -- the current is like a conveyor belt that gets "turned on" everywhere at once in the circuit as the emf (voltage, potential difference) is established through it.
So electrons in the wire immediately start shifting as soon as there is a PD. The symbolic electron leaving the plate due to the PD is NOT the current.

First, make sure you understand one fact about a sinusoidal voltage source: the voltage changes most rapidly through the zero-crossing. (Look at the zero-crossings of a sine wave if you don't immediately recognize this fact.)
I'm trying to piece together what I know here. What I'm drawing for this is the AC voltage is described by the sin wave, and its changing most rapidly at its zero points, because at these points its rate of change function (the cosine) is maximized. Is this right? Tear this one up please.

You also hopefully know that the current into a capacitor is proportional to the rate of change of the voltage across it. In other words, the current into the capacitor is largest right when the voltage is zero. The current into the capacitor is largest at the zero-crossings of the source.
I know this as [itex] Q = CV(t) [/tex], where V(t) is a sine wave in this case. The charge stored in a capacitor is given by the capacitance and the voltage difference. (I think the VD is that of the plates, which is in turn due to the V applied at the source, correct this please)
I can derive what your saying by taking the time derivative to get the current

$$I = C\frac{dV}{dt}$$.

The current flowing into the capacitor would be out of phase by 90 degrees because of the identity

[itex] cos(x) = sin(90-x)[/tex] in degrees. (This is exactly why, right?)

chroot
Staff Emeritus
Gold Member
whozum,

Sounds like you nailed it. - Warren

Ok, thanks alot. I'd like to elaborate a bit on that last part because I dont understand it completely. Say our circuit exists such that an AC signal source is connected to a capacitor with a switch in the circuit. At t = t0 the swithc is flipped, and the E field within te circuit is almost instantly established (at v = ~c). This E field provides an attractive acceleration for each electron in the circuit, and in turn current begins to flow.

The rate of flow of the current is not constant, since the voltage function is not constant. In a simple AC circuit with a resistor, the current flowing is directly in phase with the voltage, and is merely scaled in magnitude by the resistance. Here, the current is also a sin wave.

This is where things start coming out of my ass.
With a capacitor however, things change because now current is not flowing uniformly with voltage. The E field while propagating through the circuit is established inside the capacitor, and electrons begin piling up on one end of the capacitor soon after. They can not cross because the E field strength is smaller than the dielectric breakdown point. This pileup of charge on one end of the capacitor increases the voltage difference within the capacitor, which was initially zero at t = t0. The voltage increases asymptotically until it is fully charged, and the potential differnce between the capacitor's plates is the same as that of the source voltage.

Finding the time derivative of the charging of the capacitor, we find that the current flowing into the capacitor is initially I = V/R (resistance, or impedance?), and drops while the VD is established between the plates. Current is NOT flowing out of the capacitor since the charge is piled up on the plates. Sicne the source voltage is a sin wave, its time derivative would be a cosin wave of equivalent amplitude, provided no resistance in the circuit (ideal).

Sorry to be taking up so much of your time.

chroot
Staff Emeritus
Gold Member
I'm honestly not really sure if I can decipher clear, direct statement out of the second half of your last post to judge whether or not you're fully grasping the concepts.

The generalized Ohm's law I = V/Z is always true, even for capacitors and inductors -- it just happens that Z is purely imaginary for those devices, so Z is a proportionality constant not of amplitude, but of phase.

- Warren

True/False

With a capacitor do things change because now current is not flowing uniformly with voltage?

Is the E field propagating through the circuit established inside the capacitor immediately, or build up with time?

Electrons begin piling up on one end of the capacitor soon after.

They can not cross because the E field strength is smaller than the dielectric breakdown point.

This pileup of charge on one end of the capacitor increases the voltage difference within the capacitor, which was initially zero at t = t0.

The voltage increases asymptotically until it is fully charged, and the potential differnce between the capacitor's plates is the same as that of the source voltage.

The current as measured at any point is V/Z, and decreases as the capacitor is charged. This is because the capacitor has developed a potential equal to that of the battery, and there is nothing accelerating electrons (as t tends to infinity)

chroot
Staff Emeritus
Gold Member
whozum said:
With a capacitor do things change because now current is not flowing uniformly with voltage?
I don't know what "things" means.
Is the E field propagating through the circuit established inside the capacitor immediately, or build up with time?
The electric field in the capacitor depends on the charge stored within it. Since the charge builds up over time, so does the electric field.
Electrons begin piling up on one end of the capacitor soon after.
Soon after what?
They can not cross because the E field strength is smaller than the dielectric breakdown point.
Right.
This pileup of charge on one end of the capacitor increases the voltage difference within the capacitor, which was initially zero at t = t0.
Right.
The voltage increases asymptotically until it is fully charged, and the potential differnce between the capacitor's plates is the same as that of the source voltage.
Right.
The current as measured at any point is V/Z, and decreases as the capacitor is charged. This is because the capacitor has developed a potential equal to that of the battery, and there is nothing accelerating electrons (as t tends to infinity)
Right.

- Warren

Soon after what? Soon after the E field is established.

With a capacitor do things change because now current is not flowing uniformly with voltage? This was in relation to the resistor paragraph.

I've sent you a PM. Thanks for your help.