# Why don't static shocks kill you?

1. ### bhoot

6
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? Many people say that it doesn't have enough amps, but lets look at Ohms law: V = IR. I measured the resistance from one hand to the other as being 750,000 ohms in my body. That means that there are 100000 Volts/750000 Ohms = 0.13 amps going through your body. It only takes 65 milliamps to stop your heart. There have been many class demonstrations where a student holds a Van de Graaff generator and touches someone else to pass hundreds of thousands of volts through their body. Why does't this kill you?

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3. ### rollcast

418
Because electricity takes the path of least resistance to get to ground and that doesn't usually involve passing into the body.

4. ### Dickfore

Also, even if it goes through the body, it still completes a loop through air (do you actually touch the Van der Graff generator, or simply approach your finger)? Furthermore, how did you find that the resistance of your body is only 750 kΩ?

5. ### Integral

7,341
Staff Emeritus
Typically Van de Graffs only generate a few micro Amps. Way below a fatal current which is usually given as 10 to 50 milli amps.

6. ### bhoot

6
There is no path for the electricity to go but through your body. You actually touch the generator (while standing on an electrical insulator so that electricity can't pass through your feet to the ground), and then touch another person. I got 750,000 ohms by measuring my body with a multimeter.

7. ### bhoot

6
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.

### Staff: Mentor

V=ir and the numbers you are trying to plug in only apply to a steady state. The van de graf just doesn't store enough energy: it gets dissipated long before affecting your heart.

9. ### bhoot

6
Ok, I suppose that explains it. Where exactly does the energy dissipate to?

10. ### Integral

7,341
Staff Emeritus
No power supply exists which can produce an unlimited current every supply is current limited in some way. A Van de Graff, or a door knob for that matter, can only produce a "few" electrons for current flow. It may be at a very high potential but the current flow is limited by the electrons available. I quoted few because of the huge numbers involved in these processes.

### Staff: Mentor

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.

12. ### bhoot

6
This may be slightly off topic from the static electricity, but what about transformers then? Would 120 volts from house power taken up to as 100,000 volts with a transformer be able to supply enough current?

13. ### Antiphon

There is more than enough current in a carpet shock to kill you. But there isn't enough charge (not enough capacitance) to sustain the otherwise lethal current.

14. ### DrZoidberg

406
The resistance from one hand to the other is about 1,000 Ohms. 750,000 Ohms is the resistance of your skin. But when the voltage is high enough, the skin looses it's resistance and you are left with the 1,000 Ohms of internal resistance. A Van de Graaff generator can not produce much current. Let's say it's current is 10 µA. V = I*R. 10µA * 1,000 Ohms = 10 mV. That's the voltage that drops over your body. 100,000 V is what you get when the generator is insulated from ground. When the generator is grounded, whether it's through your body or through a wire, the voltage drops to nearly nothing. Btw. a Van de Graaff generator is a constant current source. It's output current is always the same independent of what resistance is attached to it.

15. ### Containment

18
It's sorta like this but not exactly... If I toss 1,000 pillows at you very softly you probably will just think it's a pillow fight but if I toss 2 or 3 pillows at you fired from a cannon at 1,000 mph your brains will go splat. Same deal with electricity the pillows are basically the voltage and the amperage is basically the speed they are moving at. So as long as the amperage is low enough below like 1 amp you probably wont die from the shock alone.

16. ### mrspeedybob

731
This is incorrect.
1 amp is a large amount of current when considering electric shock. 1/20 of that could kill you.

17. ### Containment

18
I wasn't trying to be super accurate as obviously even a low amperage shock could kill you given it hits you in the right spot or for long enough. You obviously don't want to be messing around with anything electrical if you don't understand it. Also 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. With that said humans are fairly resistant to electrical shocks for example farmers have to bring in cows from thunderstorms not because of rain but because if a bolt of lightning hits the ground the electricity passing through the ground could be enough to kill them while a human would probably be fine.

18. ### bhoot

6
So what exactly does it mean that when you bring up voltage with a transformer, you lose current?

19. ### Integral

7,341
Staff Emeritus
A transformer that steps up voltage will step down current in the same ratio. If secondary voltage is 10x the primary voltage the secondary current will be .1x the primary current.

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

20. ### Dickfore

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:
$$\begin{array}{l} \mathcal{E}(t) - L_1 \, \frac{d I_1}{d t} + M \, \frac{d I_2}{d t} = 0 \\ -L_2 \, \frac{d I_2}{d t} + M \, \frac{d I_1}{d t} = R \, I_2 \end{array}$$

Assuming harmonic time dependence, these differential equations become ordinary algebraic equations:
$$\begin{array}{l} -i \, \omega \, L_1 \, I_1 + i \, \omega \, M \, I_2 = \mathcal{E} \\ -i \, \omega \, M \, I_1 + (i \, \omega \, L_2 - R) \, I_2 = 0 \end{array}$$

The determinant of this system is:
$$\Delta = \left\vert \begin{array}{cc} -i \, \omega \, L_1 & i \, \omega \, M \\ -i \, \omega \, M & i \, \omega \, L_2 - R \end{array} \right\vert = \omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega$$
and the auxiliary determinants are:
$$\Delta_1 = \left\vert \begin{array}{cc} \mathcal{E} & i \, \omega \, M \\ 0 & i \, \omega \, L_2 - R \end{array} \right\vert = (i \, \omega \, L_2 - R) \, \mathcal{E}$$

$$\Delta_2 = \left\vert \begin{array}{cc} -i \, \omega \, L_1 & \mathcal{E} \\ -i \, \omega \, M & 0 \end{array} \right\vert = i \, \omega \, M \, \mathcal{E}$$
so the currents are:
$$\begin{array}{l} I_1 = \frac{(i \, \omega \, L_2 - R) \, \mathcal{E}}{\omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega} \\ I_2 = \frac{i \, \omega \, M \, \mathcal{E}}{\omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega} \end{array}$$

For ideal transformers:
$$L_1 \propto N^2_1, L_2 \propto N^2_2, \, M \propto N_1 \, N_2 \Rightarrow L_1 \, L_2 = M^2$$
and the above expressions simplify to:
$$\begin{array}{l} I_1 = \frac{\mathcal{E}}{R} \, \left( \frac{N^2_2}{N^2_1} + i \, \frac{R}{\omega \, L_1} \right) \\ I_2 = \frac{\mathcal{E}}{R} \, \frac{N_2}{N_1} \end{array}$$
As you may see, the real parts of these currents are in a ratio:
$$\frac{I'_1}{I'_2} = \frac{N_2}{N_1}$$
The voltage drop, on the other hand, is given by:
$$V_1 = L_1 \, \frac{dI_1}{d t} - M \, \frac{d I_2}{d t} \rightarrow -i \, \omega \, (L_1 \, I_1 - M \, I_2)$$
$$V_1 = \mathcal{E}$$
$$V_2 = L_2 \, \frac{d I_2}{d t} - M \, \frac{d I_1}{d t} \rightarrow - i \omega \, (L_2 \, I_2 - M \, I_1)$$
$$V_2 = \mathcal{E} \, \frac{N_2}{N_1}$$
As you can see, the ratio of the voltage drops is:
$$\frac{V_1}{V_2} = \frac{N_1}{N_2}$$
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:
$$V_R = R \, I_2 = \mathcal{E} \, \frac{N_2}{N_1}$$

21. ### Andy Resnick

5,744
As other people have noted, "It's the volts that jolt, but the mills that kill."

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: