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Integral

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Can you explain how the amperage is low even though the volts are high? I have some numbers in the original post, 100 kV for the generator and 750,00 ohms for the body. I get .13 amps. with V = IR.

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russ_watters

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Ok, I suppose that explains it. Where exactly does the energy dissipate to?

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Integral

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russ_watters

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Well that's just it: there is very little energy to dissipate in the first place! Almost all woulg get dissipated right at the site of the shock.Ok, I suppose that explains it. Where exactly does the energy dissipate to?

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This is incorrect.

1 amp is a large amount of current when considering electric shock. 1/20 of that could kill you.

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So what exactly does it mean that when you bring up voltage with a transformer, you lose current?

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Integral

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Power in must equal power out and power = Current x Voltage.

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A transformer works according to a principle of mutual inductance. A changing current in the primary coil induces an emf in the secondary coil, and vice versa. There is still self inductance present, creating a "back emf" from the changing current in its own coil. If you attach a source of emf in the primary and an active load (resistor) in the secondary, then the equations become:So what exactly does it mean that when you bring up voltage with a transformer, you lose current?

[tex]

\begin{array}{l}

\mathcal{E}(t) - L_1 \, \frac{d I_1}{d t} + M \, \frac{d I_2}{d t} = 0 \\

-L_2 \, \frac{d I_2}{d t} + M \, \frac{d I_1}{d t} = R \, I_2

\end{array}

[/tex]

Assuming harmonic time dependence, these differential equations become ordinary algebraic equations:

[tex]

\begin{array}{l}

-i \, \omega \, L_1 \, I_1 + i \, \omega \, M \, I_2 = \mathcal{E} \\

-i \, \omega \, M \, I_1 + (i \, \omega \, L_2 - R) \, I_2 = 0

\end{array}

[/tex]

The determinant of this system is:

[tex]

\Delta = \left\vert \begin{array}{cc}

-i \, \omega \, L_1 & i \, \omega \, M \\

-i \, \omega \, M & i \, \omega \, L_2 - R

\end{array} \right\vert = \omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega

[/tex]

and the auxiliary determinants are:

[tex]

\Delta_1 = \left\vert \begin{array}{cc}

\mathcal{E} & i \, \omega \, M \\

0 & i \, \omega \, L_2 - R

\end{array} \right\vert = (i \, \omega \, L_2 - R) \, \mathcal{E}

[/tex]

[tex]

\Delta_2 = \left\vert \begin{array}{cc}

-i \, \omega \, L_1 & \mathcal{E} \\

-i \, \omega \, M & 0

\end{array} \right\vert = i \, \omega \, M \, \mathcal{E}

[/tex]

so the currents are:

[tex]

\begin{array}{l}

I_1 = \frac{(i \, \omega \, L_2 - R) \, \mathcal{E}}{\omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega} \\

I_2 = \frac{i \, \omega \, M \, \mathcal{E}}{\omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega}

\end{array}

[/tex]

For ideal transformers:

[tex]

L_1 \propto N^2_1, L_2 \propto N^2_2, \, M \propto N_1 \, N_2 \Rightarrow L_1 \, L_2 = M^2

[/tex]

and the above expressions simplify to:

[tex]

\begin{array}{l}

I_1 = \frac{\mathcal{E}}{R} \, \left( \frac{N^2_2}{N^2_1} + i \, \frac{R}{\omega \, L_1} \right) \\

I_2 = \frac{\mathcal{E}}{R} \, \frac{N_2}{N_1}

\end{array}

[/tex]

As you may see, the real parts of these currents are in a ratio:

[tex]

\frac{I'_1}{I'_2} = \frac{N_2}{N_1}

[/tex]

The voltage drop, on the other hand, is given by:

[tex]

V_1 = L_1 \, \frac{dI_1}{d t} - M \, \frac{d I_2}{d t} \rightarrow -i \, \omega \, (L_1 \, I_1 - M \, I_2)

[/tex]

[tex]

V_1 = \mathcal{E}

[/tex]

[tex]

V_2 = L_2 \, \frac{d I_2}{d t} - M \, \frac{d I_1}{d t} \rightarrow - i \omega \, (L_2 \, I_2 - M \, I_1)

[/tex]

[tex]

V_2 = \mathcal{E} \, \frac{N_2}{N_1}

[/tex]

As you can see, the ratio of the voltage drops is:

[tex]

\frac{V_1}{V_2} = \frac{N_1}{N_2}

[/tex]

which is an inverse relation to the one obeyed by the ratio of the active parts of the currents.

The voltage drop on the resistor load is also:

[tex]

V_R = R \, I_2 = \mathcal{E} \, \frac{N_2}{N_1}

[/tex]

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As other people have noted, "It's the volts that jolt, but the mills that kill."I know that lightning can, but look at the following example: A Van de Graaff generator can produce lots of volts. Lats say 100 KV for our purposes. Why doesn't this voltage kill you?

In terms of applying Ohm's law to yourself, your skin is a very good insulator- the current tends to flow along the surface of your skin via the highly conducting sweat. If the current were to penetrate, then you'd have a major problem.

Even so, the human body is not a simple insulator. Electrical models of the body are here:

http://en.wikipedia.org/wiki/Human-body_model

And a discussion over the relative dangers of DC and AC here:

http://www.allaboutcircuits.com/vol_1/chpt_3/2.html

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Also, do we know that the human body obeys Ohm's Law? Not everything does.

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Office_Shredder

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I was under the impression this only happens in moviesAlso something to consider is that while a dc shock may not kill you electrically when you get tossed 6 feet away into the wall you might die

- #23

sophiecentaur

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The maximum 'shock' charge that a home 'power breaker' will cut out at is more than 1000 times that value and even that is a pretty conservative level. This is AC, of course, which is a bit different in practice but it gives you an idea.

For DC shocks, if the capacitance is much greater than a VDGG - like the Leyden Jars on a Whimshurst machine, which was what we had at School - you can get a really nasty / lethal shock even when the voltage generated is only a few tens of kV. That's why they don't have any in modern Schools (bah, no fun).

Actually, the resistance of your body is not very relevant when dealing with high static voltages and low Capacitances. The resistance just determines the actual current - which will affect the time taken for the discharge ( shock) to take but not the amount of actual charge.

Resistance does count (a lot) when you are dealing with shocks from 'powered equipment' which usually involves only moderate voltages of hundreds or thousands of volts. Above that, it's really Russian Roulette and depends upon things like the frequency content of the spark and the geometry of your body and surroundings to determine where the current happens to travel over or through you. People have been known to survive incredible lightning strikes, involving thousands of Amps - but clearly not actually 'through' them.

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nsaspook

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I got hit once by a flying helicopters cargo skid before it was discharged. The sudden contraction of my back and leg lifted me off the ground and back a few feet. My body hurt for a week.I was under the impression this only happens in movies

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sophiecentaur

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