14
(a) "How does the circuit "know" that it has to maintain a constant current?"
If the current (rate of flow of charge) wasn't the same all round the circuit, then electric charge would be piling up at some point or points. This couldn't go on happening for long because the piled-up charge (negative, let's say) would prevent (by repulsion) ...
11
I believe you've got the title question backwards, and are ignoring the definition of resistance. Let's examine resistance first.
Resistance for a simple resistor is a ratio of the energy absorption per charge to the rate of charge flow, i.e., potential difference/current. But you know that. The conceptual difference, however, is that the resistor doesn't &...
6
You can see that this is a very small capacitor and rated 100μF.
It's not really so small. 100 uF surface-mount capacitors are available in smaller packages than that. As small as 2.0 x 1.25 x 1.45 mm, for example, but with fairly low voltage rating at that size. Or 3.2 x 1.6 x 1.9 mm with the same 10 V rating as the one you have.
I actually don't know the ...
5
In simple terms, going around a closed loop Kirchhoff tells you that the sum of the potential differences is zero.
For example, suppose a simple series circuit with three nodes $a,\,b $ and $c$, then $V_{\rm a \to b}+V_{\rm b \to c}+V_{\rm c \to a}=0$.
Multiply each term by the current $I\Rightarrow IV_{\rm a \to b}+IV_{\rm b \to c}+IV_{\rm c \to a}=0$ ...
5
A vector field $\boldsymbol{A}$ is conservative if the integral
$$\int_{\Gamma}\boldsymbol{A}\cdot d\boldsymbol{l}\tag{1}$$
is path independent i.e. it does not depend on $\Gamma$.
$(1)$ is equivalent to
$$\oint\boldsymbol{A}\cdot d\boldsymbol{l}=0\tag{2}$$
for any closed path.
Now, electrostatic energy difference between to points $A$ and $B$ is given by
$$\...
5
Current flowing in a closed circuit is just a special sub-case required for continuous operation of electronic circuits. In general, current and charges do not need a closed circuit to flow. Think about things like static charges, lightning bolts, and antennas. It's just that if the circuit is not closed then charge accumulates and eventually cancels out the ...
4
But if it goes all the way to zero, would it turn into effectively a series circuit and cause the [total] resistance to jump back up?
should read
. . . . if it goes all the way to zero, it [would] turn into effectively a short circuit and cause the [total] resistance to become zero.
4
what's stopping the current from rising to infinite
Assuming no driving voltage, typically the capacitor starts with some finite amount of charge, or a finite amount of current is in the circuit. This is analogous to a mass-spring system that starts at a finite displacement or a finite initial velocity.
And if resistance is provided by the inductor by self ...
4
For the formula to work, the conductor must be a (right) general cylinder. That is, its cross section must be the same at all points along its length.
Then the cross-sectional area is the same no matter which location along the length you choose to measure it at.
The cross sectional area is simply the actual cross-sectional area, regardless of how complex ...
4
When the voltage over a circuit changes, the change will propagate through the circuit. During a very short time, the voltages and currents will fluctuate, and the current can even differ along a conductor. But quickly the voltages and currents will find an equilibrium according to the different parameters of the circuit (like resistances and voltages of ...
4
The key is to think about the electric field, not the electrons. It is true that the electrons do move (slowly) but it is the electric field across the different parts of the circuit that makes this happen - and changes in the electric field propagate almost instantaneously.
If there were a potential difference between, e.g. the yellow and blue paths in your ...
3
If you are looking for an more intuitive understanding then maybe look at 'cause and effect'
A circuit with 0 voltage at any point evidently has no current flowing.
A potential difference provides a 'push' (the water pressure analogy DOES work here) to make the current flow.
It seems intuitive that the harder you push the more will flow. Ohms law simply ...
3
For many physics problems the way to solve them is to get the right level of abstraction. You have drifted into the wrong level.
You need to be careful about comparing currents in wires to motion of particles affected by potential differences on static charges. Your diagram shows a test charge being moved from infinity towards a fixed charge. Electric ...
3
ASSUMING it is properly connected (no faults and bonded to ground at
mains) why is it I don't get a shock? And more-over to my
understanding of AC power, how is the ground cable then not carrying
at least some current?
By bonding the neutral conductor to ground at the service panel, the neutral conductor is close to ground potential. I say "close" ...
3
Current always flows in a loop. In this case, from the battery, through the wrench, and back to the battery. The experimenter's body does not provide a path for current in that loop. It would be different if the experimenter left the wrench in the toolbox and simply grabbed the '+' terminal with one hand and the '-' terminal with the other. In that case, ...
3
You are referring to transmission line effects. Things such as reflections, oscillations, ringing, and voltage spikes that occur whenever there are changes/transients in the circuit. This includes things like connecting power up to an otherwise DC steady state circuit.
These unsteady state signals bouncing back and forth between components of the circuit is ...
3
For simplicity, assume that the loop's circumference is small compared to the wavelength, consequently the current in the loop is essentially constant along the perimeter at any instant. That loop will not respond to the electric field because it is in the plane of the loop and therefore there will be as much force moving the charges in one direction along ...
2
An oscillating dipole with a large enough dipole moment would cause the bulb to light up. If the resistor is just a resistor with short conductive leads on each end then what happens is the original dipole induces an induced oscillating dipole moment in the resistor. This induce dipole moment results in an electric field pointing into the resistor, a voltage ...
2
You can consider that wire as being one big node. I suggest you redraw the circuit in such a way that you replace the encircled wire with a node.
I don't understand why the bottom wire can be ignored.
It is not short circuit since it does not directly connect two terminals of the same element.
To solve for the output voltage $v_o$ I suggest you use the ...
2
To answer your question it is necessary to explain how a transformer works. Also you have to understand the complex notation for currents and voltage. I am not going to explain it, if you don't understand it, please check some physics textbooks.
Both coils of the transformer are wound around the same core. Let us say that the primary goes around the core $...
2
If there is no resistance in the circuit, the rate of change of current in the circuit does not change, as explained here: inductor back EMF.
Thus, the decrease in the magnitude of the rate of change of current as time progresses must relate to there being resistance in the circuit.
In the circuit, you have $\mathcal E_{\rm Battery} - V_{\rm resistance} = \...
2
A simple intuition is that when we increase the area of cross section, more number of electrons can pass through the area as compared to initial wire. More electrons results in more charge crossing the cross section, thus more current.
Now let's look at it with some Math:
Resistance of a wire is given by :
$$R = \rho \frac{l}{A}$$
$R$ = resistance of the ...
2
The current will still be constant, otherwise there would be a violation of conservation of charge.
What will be different is the electric field in the resistor. For a uniform resistor the electric field is uniform in the resistor with a magnitude of
$$E=\frac{V}{d}$$
Where $V$ is the voltage drop across the entire resistor and $d$ is the length of the ...
2
If you're driving a car you are using up energy to do so, in the sense that you are burning calories to move your body to control the vehicle. But the gas burned in the engine is what moves the car. You use energy to control a device that will use a lot more energy.
A CPU is just many many electric circuits, inside of all of those circuits charge flows ...
2
Circuit theory is inherently non-relativistic. You simply cannot use circuit theory to analyze a circuit moving at relativistic velocity. When a circuit is moving relativistically you need to use Maxwell’s equations to analyze it and cannot take the usual shortcuts afforded by circuit theory.
The issue is the following. Circuit theory rests on three ...
2
Power lines have slightly higher resistance for AC than for DC due to the skin effect.
The wiki article has extensive information on your question, including the AC vs. DC resistance of round wires. In practice, as power wires are stranded, the skin effect losses can be kept very small.
Why is AC still used? You also have to factor in power generation ...
2
The EMF gives energy to a charge and the charge transports the energy to somewhere else.
This is not usually the case. In most cases the energy is transported in the fields, not in the charges.
For example, consider an ordinary light switch and light bulb. The charges move at rates of a couple of mm/s, so if the light bulb were located a couple of meters ...
2
Current is not just the speed of the flow. It is the amount of charge passing a given point per unit time. If it is not the same amount all around the loop forming the circuit then that means the charge must be building up somewhere, and falling away somewhere else. This cannot go on forever so it can't be happening if the situation is in a steady state with ...
1
Kirchhoff's voltage law is a consequence of the scalar nature of Coulombs law and the conservative nature of the Coulomb force. The Coulomb potential only depends on position, not on the path connecting positions. It is valid only if the current is stationary. And yes it is also consistent with energy conservation.
1
I interpret the question as asking to a resistor of gradually different resistances across its geometry. In other words, the resistance is constant (over time) but differes from point to point along its length.
Will the current be the same at all points throughout your resistor?
Yes.
Here are two arguments for that conclusion:
The resistance perspective
A ...
Only top voted, non community-wiki answers of a minimum length are eligible