# Why don't static shocks kill you?

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

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?

Integral
Staff Emeritus
Gold Member
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.

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.

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.
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.

russ_watters
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.

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

Integral
Staff Emeritus
Gold Member
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.

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

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?

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.

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.

Containment
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.

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

Containment
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.

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

Integral
Staff Emeritus
Gold Member
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.

So what exactly does it mean that when you bring up voltage with a transformer, you lose current?
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:
$$V_R = R \, I_2 = \mathcal{E} \, \frac{N_2}{N_1}$$

Andy Resnick
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?
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:

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

Office_Shredder
Staff Emeritus
Gold Member
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
I was under the impression this only happens in movies

sophiecentaur
Gold Member
If you charge the ball of a Van der Graaf generator to 500kV, there is very little charge stored. The capacity of a small VDGG ball is only about 20pF and Q = CV, so we're talking in terms of only about one microCoulomb (give or take) - not enough to do anything more than attract your attention!
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.

nsaspook