81
A birds legs are pretty close together. An electrical transmission wire has very little resistance.
This means that the voltage as a function of distance barely changes. So the voltage difference between two birds feet is essentially 0, because the potential on each foot is practically the same. The potential difference between the wire and the ground ...
37
Let me say in advance: You are perfectly right! The wire is a resistor, the bird is a resistor, and a bird standing on a wire with both feet down is indeed a parallel resistor to the wire. This means indeed that current flows through the bird. It is just not much, because the wire is — by design! — a very bad resistor, and birdie is by comparison ...
23
Electric potential is a difference between two points, and considering that the wire the bird is standing on has little resistance, the potential difference would be negligible. That means when the bird is standing with both feet on the wire, the potential difference between its two feet is minuscule and with its own high resistance, will most definitely not ...
10
Let me first note that the voltage itself does not have any physical effect - the harm comes from the electric current. This happens in two ways:
via the Joule heat when high current runs through the body (this requires currents of several Amperes, that we rarely encounter in the everyday life, but it is relevant for the bird)
via triggering the heart ...
7
Qualitatively: the low electrical resistance of power lines means that there is negligible difference in electrical potential between two closely spaced points on the line. As a result, barely any current will flow through a (resistive) bird on the line, and the vast majority of current will run through the (conductive) line as normal.
Quantitatively: the ...
5
While one can specify a voltage at a point, it must always be with respect to another point, sometimes referred to as common, ground or reference point where the voltage is arbitrarily assigned a value of zero. The zero voltage point is typically chosen at the negative terminal of the voltage source.
For the simple circuit below, I have labeled points A, B,...
5
The size of the voltage in the wire is not what determines the current through the bird. It is the difference in the voltage between the bird's feet. If the wire is charged by a 600V source, the voltage between the birds legs is not 600V. The actual voltage difference is very small. The birds legs are touching two points very close to each other on the wire. ...
3
in Ohm's law we are measuring voltage at a point so work done should be very small since there is almost no distance
This is the key mistake. In Ohm’s law the voltage is the voltage across the resistor. In other words, the voltage in Ohm’s law is measured at two points. One point is always the point where the current is entering the resistor and the other ...
3
Am I missing something?
Yes, you are missing what is held constant in each case.
Ohm's Law: I=V/R. Increasing voltage increases current.
Should read “Increasing voltage increases current for a fixed resistance.”
Power Law: P =𝑉∗𝐼. Increasing voltage decreases current.
Should read “Increasing voltage decreases current for a fixed power.”
The two ...
2
Lets make clear from the beginning that Physics theories use mathematics as a tool, imposing extra axioms called postulates, principles,laws so that those mathematical solutions are picked that fit measurements and observations and also, very important, predict new phenomena which become validated.
Electricity and magnetism were slowly defined and measured ...
2
According to diode theory, there is no such thing as a turn-on voltage. The current-voltage characteristic of the junction is given by
$$I = I_0 (e^{\frac{qV}{kT}}-1),$$
where $kT/q \approx 25$ mV at room temperature. At voltages larger than 100 mV, the unity term can be ignored.
In this equation, $I_0$ contains the dependence on band gap. It also depends ...
2
The shorter wavelength the more voltage required because the photon that is needed to interact with the semiconductor has to have more energy. So colors like blue will require more voltage than colors such as red. Different colors use different materials and the materials that produce the shorter wavelengths require more energy to move the electron to a ...
2
increasing the gas pressure increases its density, which decreases the mean free path length of moving ions. This limits the distance an ion can travel before it gets deionized in a collision, and increases the amount of electric field strength required to achieve a breakdown cascade.
2
As the others say, you are implicitly holding one factor constant and seeing how the others are related but a different constant factor in each case.
A real world example may help. Consider an old style filament light bulb. It is intended to be used in a $220V$ country and consume $110W$. So, it should draw $0.5A$ and needs a resistance of $440 \Omega$. ...
2
Sometimes notation may induce misunderstandings. It is true that in many formulas voltage appears as $\Delta V$, making clear that it refers to the difference of voltage between two points. However, arrived at Ohnm's law, it is usual to find it expressed as
$$
I=\frac{V}{R},$$
thus inducing to think that only voltage at one point is present.
Actually in ...
2
Analogy
Think of water running through a pipework: it may run faster where the pipes are narrower, and slower in the wider parts, but the quantity of water which enters is the same as the quantity of water that exits. There may be temporary decompressions here and there, but the water eventually has to come out, unless there is a leak or a reservoir ...
1
The current in the rightmost 5 Ohm resistor has no influence on the current in the 10 Ohm resistor and vice versa because the voltage across the 5 Ohm resistor is fixed at 10 volts. It can therefore be ignored.
1
I have a couple of issues with your analysis.
The first is your treatment of air like a battery supply. In order for the air to supply current to the body, it would need to become ionized so that free electrons are available. That requires the air to undergo dielectric breakdown. The breakdown voltage for air depends on a number of variables including the ...
1
definition of voltage as work done moving unit of electric charge between two points in electric field
That's almost okay, but what's missing is the concept of electrical potential. And technically, you should say "work per unit charge" to get the units correct (volt = joule per coulomb). That's often not discussed in a lower level (high school) first ...
1
The wire will equalise the potential between the contact points on the insulators. It cannot do any more than that, because the insulators will not conduct. Because they cannot conduct, charge cannot flow from other parts of the insulators, and hence they will stay at their original potential.
EDIT
I should add that there is no such thing as a perfect ...
1
Here's how I understand what you are saying. Correct me if I'm wrong. A circuit might be rated for 25 A, but the appliance uses only 2 A. That leaves 23 A left over. You wonder why those left over amps don't heat things up.
A resistor, an appliance, or anything else, heats up only when current is flowing through it. Massive amps do not come from the ...
1
The issue is this:
This would mean Chloride is flowing from a Low potential to a High Potential(When the natural order of things is High to Low)
Positive charges move from higher to lower potential. However, since chloride ions are negatively charged, they actually want to move from lower to higher potential. This is because they want to move to lower ...
1
The potential difference between two points is equal to the work done by an external force in moving unit positive charge between those two points.
So for your top diagram with the positive charge starting on the left, potential $V$, and moving to the right, potential $0$, the work done by an external force on a charge of $+1\,\rm C$ is $-12\rm \,J$.
...
1
The key point here is that a resistor can handle both causal forms of ohm's law: If we assert a fixed voltage as an input the resistor returns a current as an output, or if we assert a fixed current as an input, the resistor returns a voltage drop as an output.
For example, think of an ideal current source as a positive displacement water pump that runs at ...
1
Here's one way to think about it.
Consider the case that one cell is connected across a resistor. Stipulate that the cell has a capacity of 1 amp-hour and that the cell delivers 1 amp of current to the resistor when first connected.
You should expect that the cell will discharge in roughly 1 hour.
Now, disconnect the discharged cell, and connect to the ...
1
My textbook says, that across a battery or a cell there should be a
potential drop equal to its e.m.f.
To be clear, you textbook is referring the potential measured across the battery terminals without any circuit connected to the terminals, i.e., the no load or open circuit potential across the terminals.
Now I've also learnt from sites that the ...
1
across a battery or a cell there should be a potential drop equal to its e.m.f.
Sometimes called the "open circuit voltage", this potential only exists when there is no current flowing through the battery. Internal resistance within the battery will reduce that potential as current increases.
I've also learnt from sites that the potential at all points ...
1
It has nothing to do with $r’$ being zero. The only place $r’$ is zero is at the origin.
When $\rho$ is spherically symmetric, it is independent of the angular coordinates, say the usual ones $\theta’$ and $\phi’$. Then, if one takes the $z$-axis along $\mathbf r$ so that the angle $\alpha$ between $\mathbf r$ and $\mathbf{r}’$ is just $\theta’$, the $\...
1
Because it is a potential difference, it doesn't matter.
For example: "I'm 10cm taller than my dad" and "my dad is 10cm smaller than me" contain the same information, just expressed differently.
Similarly, suppose I measure $V_{AB}$ and say, "$V_{AB}$ is 5V", and you say "actually it is -5V", we still agree on the actual quantity, but just disagree on how ...
1
The physics of both equations is the same. Look at the second equality in each one. Both agree on $V_B-V_A$. It’s only ambiguous notations $V_{AB}$ and $\Delta V$ that are confusing.
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