# Tag Info

38

Which explanation is more correct? The answer to the second question you cite is the best one. In order to be "electrocuted" a non-trivial amount of current must flow through the body. The amount of current that flow is a function of the impedance of the bird and the voltage difference between the two contact points. The second point is crucial ...

14

A human's skin resistance is not 300 Ohm. But an actual 200,000V is enough to kill you if it actually reaches you, whatever the skin resistance is. However, the source's output impedance and energy behind the source might prevent it from actually reaching you at full voltage. The 200,000V was measured when zero current was flowing. If the source's output ...

12

If, say, 3 charges enter each second then also 3 charges must leave each second for a steady state current. If more leave than enter each second, then where would the extra leaving charges come from? This is not possible. If fewer leave than enter each second, then some charges are staying behind within the conductor. Over time the net charge in the ...

10

I believe your question is about how a current can remain constant after passing through a conductor. The water analogy introduced by Roger Vadim reminds me of another explanation. A kid asks HC Verma, a renowned Indian physicist, Why does current not decrease on passing through a resistance. HC Verma's explanation is in some sense a "Proof by ...

10

Edit due to comment: This is a partial answer, it addresses the title of the question: Are the "bird sitting on a live wire" answers wrong? which according to OP suggest that its actually the no grounding that prevents the bird from getting fried. (Along with having both feet in effectively the same place) Even if one treats the bird as a ...

8

I think your question seems to imply that you think that charges at one end of the wire are "fired" through the wire with some initial kinetic energy and and the resistance slows it down. The electric field inside a wire is constant so at each point in the wire charges experience a force Eq the resistive forces inside the wire are proportional to ...

6

Nobody in the previous answers seemed to have mentioned anything about the alternating nature of the current and impedances etc generated in the bird's body due to that. This is not correct, as my own answer to one of the cited questions explicitly mentions that ac and dc currents differ in their harmful effect. The danger of the ac current is in causing ...

4

The $V$ in $VI$ and the $V$ in $V^2/R$ are indeed the same thing. Your confusion comes because there is more than one $V$ and $I$ and $R$ in a typical circuit. In your supply cables example there is a voltage across the cable itself, a voltage across the load, and the source voltage. None of them is equal to the others. If you want to apply $V^2/R$ for the ...

4

The charge conservation (often expressed by the continuity equation, $\partial_t\rho + \nabla\cdot\mathbf{j}=0$ means that, the difference between the charge entering the conductor and the charge exiting it, is accumulated as charge within this conductor. So in your scenario, the charge of the conductor should grow to infinity. What is misleading here is ...

3

But, my question is: How the resistor reduces current from the starting of the path of current, i.e. from the point where the connecting wire is connected to the positive terminal of the battery. It doesn't. The current is the same in the resistor and the connecting wires. You can't think of current as as something where electric charge makes a journey from ...

3

For (truly) distinct nodes in a Kirchhoff circuit network, it is not appropriate to formulate the Kirchhoff current law taking in the currents from separate nodes and mixing an incomplete set from both. For the specific circuit in your question, however, the "nodes A and B", as you have described them, are not actually distinct. Any two points ...

3

Ignoring resistance, the Telegrapher's equations help to understand this question. The voltage along the circuit should behave as $\frac{d^{2}V}{dx^{2}} = u^{2}\frac{d^2V}{dt^{2}}$. This is a wave equation with speed $u = \frac{1}{\sqrt{LC}}$ with inductance-per-length L and capacitance-per-length C. This speed is the speed of light "for transmission ...

3

All reasons are relevant. If only one condition were not met, the bird would be electrocuted: If the bird were touching ground and wire, or two different phase wires, it's the only thing that stands between the electrons and a large voltage difference, and it will be well-done in no time. If the bird were standing on a good resistor, comparable to its own ...

3

Case 1 is definitely wrong. If it's so, then repeating this procedure will lead to a wire with vanishingly small length but finite resistance, which is non-sense. Case 2 it's correct that the resistance will decrease, but in a more subtle way. In general, the resistance of a material is given as (Hyperphysics): \begin{align*} R = \rho \frac{L}{A} \end{align*...

3

Using the pre-Relativity approach (valid even if unsatisfying), when you move both the magnet and the coil in the same direction you induce two emfs that cancel. There is, as you say, a non-conservative electric field due to the magnet's motion. This provides electric Lorentz forces ($q\mathbf B$) on the charge carriers in the coil. The motion of the loop ...

2

In simple terms initially there is a potential difference across the wires, current flows, due to the high resistance of air, charges will accummalate on the end of the wires, this charge accumulation creates a potential difference to counteract the batteries potential, once they are equal, no current flows

2

There is no analog to static friction in Ohm’s law. However, a diode could be somewhat analogous to a device with static friction. Unfortunately, the analogy is a little odd because the diode’s “static friction” is direction dependent. It is small in one direction and large in the other. Generally, mechanical systems are more complicated than electrical ...

2

A battery provides a force onto an electron at the +ve terminal and does work on it to move it against the electric field to the -ve terminal. This work done on the electron is stored as potential energy of the battery. This is not too bad, but you are neglecting the electrolyte. In a battery there are two electrodes. At one, an electron is consumed in the ...

2

Will electric field change ? Yes If yes then why Because you kept the potential the difference the same but increase the distance over which it is dropped. electric field is produced by the battery having some potential difference, it is only related to the battery i.e. $\Delta V = -\int \vec E.d\vec r$ or $E = -\frac{\Delta V}{\Delta r}$ $\Delta r$ ...

2

In your drawing A and B are not separate nodes, they are different points on the same node. All points connected by a single contiguous conductor are a single node. If you tried to treat them as separate nodes then you would have to consider the conductor between them to be a 0 ohm resistor. This would do nothing but add some additional equations which would ...

2

Electrons are not billiard balls. They are not isolated particles bouncing around in the wire. They interact with one another via electrostatic forces. If you put two electrons in an area, they repel each other. If you push electrons into a conductor (from a battery) without letting them out, the electrostatic forces will push the electrons apart. This ...

2

Drude model view Let us consider the simplest Drude-like model: electrons are accelerated by electric field $\mathbf{E}$ during time $\tau$ and then lose all their kinetic energy from collisions with the lattice. Between the collisions the electron velocity is governed by Newton's equation: $$m\dot{v}=-eE,$$ so that the average electron velocity is v_d = \...

2

BC is not part of any circuit. It just creates a voltage offset between the two circuits. There will only be a transient current upon connecting the 4V source.

1

1.We usually calculate the potential to infinity and not to a near charge, so You can always compare C and B. there is only the factor q between potential and potential energy, so it makes no sense to differentiate between them. a charge q gains or looses its kinetic energy if it moves in a potential difference.

1

The resistor is a component that impedes the flow of current. If, in your circuit, the resistor was replaced by a wire, then a much larger current would flow (even wires have a small and non-zero resistance). As noted, current isn't smaller in different parts of the resistor. But if the question is about how a resistor works, a (very) simple model is as ...

1

In the steady state of the circuit, the current before and after a resistor must be equal, because if there is a difference in current, there is a build up of charge inside the resistor that will go on forever. So the current next to the battery adjusts itself according to the resistance of the resistor.

1

Why is Ohm law applied in many answers and articles explaining the reason for high voltage when it is applicable for complete conversion of energy (but losses are less than 100% in mains power)? or wiki incorrectly adds this condition? The Wikipedia article on Joule heating is not incorrect in adding that condition. In the case of power distribution line ...

1

if we calculate the field between two point on a wire taking the same value of ΔV (as of battery) You cannot choose to take the potential between two points of a wire. It can be however be calculated if one knows the resistance and the current flowing through the two points. So if a current $i$ passes through the wire and the two points under consideration ...

1

Squeeze a water bottle. The pressure difference (voltage) increases across the opening, the constriction point (resistance), which causes more water particles to flow through each second (current). Note that the water current is constant. It is the same at every point. Because - and I believe this is the root of your confusion - equal current doesn't have ...

1

Basically there are 3 things to take into consideration: In order for the shock to be dangerous the current flow must be: High enough, last for a certain amount of time and pass through the dangerous areas (heart). I thinks we are talking about tens of milliamps for tens of milliseconds through the heart, but don't quote me on that. Actual body resistance: ...

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