# Is it voltage or current that shocks the most?

#### howartthou

Joined Apr 18, 2009
111
Say you have 2 circuits:

P = VI

10w = 5v x 2 amps

10w = 10v x 1 amp

Both have the same power in watts but which one would shock you more? Or would they be the same? Is the amps or the voltage that determines the intensity of the shock, or is it the power regardless of the circuit design?

#### dl324

Joined Mar 30, 2015
11,232
What kind of contact are you talking about? Dry, undamaged skin? Under most/many conditions, neither would give a shock.

#### MisterBill2

Joined Jan 23, 2018
6,694
Shocks are caused by current flowing through a body. The magnitude of that current is proportional to the voltage across the contact points. It is the current that does the damage, it is the voltage that pushes the current. Without both voltage and current there is no shock.

• MrSoftware

#### crutschow

Joined Mar 14, 2008
25,247
Perceived shock is proportional to current.
The voltage determines how much current goes through your body, depending upon you body and any other resistances in series with you and the source.
Generally voltages below about 50V will not give a significant shock to someone with normal dry skin resistance.

#### WBahn

Joined Mar 31, 2012
26,047
In addition to what the others have said, you seem to be falling prey to the common fallacy that you can choose between 5 V x 2 A and 10 V x 1 A. In nearly all situations, this is not the case. If you were to apply 5 V to something and get 2 A of current flowing, then when you change to 10 V you will almost always get more than 2 A flowing (sometimes a little bit more, sometimes twice as much, sometimes orders of magnitude more).

In general, you get to chose either the voltage or the current and the circuit will dictate the other.

#### howartthou

Joined Apr 18, 2009
111
Hmmm. I appreciate the replies, thank you, but thought I gave a good example but some of you are taking it too literally.

Lets say the circuits in question had enough voltage and current to shock you. The fact that they are the same wattage but one current is half that of the other under, obviously 2 different resistances in their load, which circuit would fry you the most given that they are the same wattage but one circuit has half the amperage of the other?

So you somehow snapped the circuit wire and put yourself in circuit. Before you do that the load in one circuit is twice the amperage of the other circuit yet they consume the same power (wattage). What circuit would fry you the most?

#### dl324

Joined Mar 30, 2015
11,232
thought I gave a good example but some of you are taking it too literally.
We went with what you gave us.
What circuit would fry you the most?
It depends. You need voltage and current to get a shock. Current will depend on voltage and skin resistance. Without a lot more information, we can't give you a meaningful answer.

#### WBahn

Joined Mar 31, 2012
26,047
Hmmm. I appreciate the replies, thank you, but thought I gave a good example but some of you are taking it too literally.

Lets say the circuits in question had enough voltage and current to shock you. The fact that they are the same wattage but one current is half that of the other under, obviously 2 different resistances in their load, which circuit would fry you the most given that they are the same wattage but one circuit has half the amperage of the other?

So you somehow snapped the circuit wire and put yourself in circuit. Before you do that the load in one circuit is twice the amperage of the other circuit yet they consume the same power (wattage). What circuit would fry you the most?
What does the wattage of the circuit have to do with anything?

As soon as you "snap the circuit wire" the original resistance that was resulting in the power draw is no longer in the circuit and is therefore immaterial.

Think of it this way. Take a 10 W light bulb and screw it into one light fixture in your house. Take a 100 W light bulb and screw it into another light fixture in your house. Now have one person go up to the 10 W bulb, remove it from the fixture, and stick their fingers in. Have another person do the same to the other fixture. Is there any reason to think that one person is going to get "fried more" based on which bulb was in the circuit before they got tangled with it? (Arguably there is, since the temperature of the 100 W fixture will likely be higher, but it would be a fairly straight matter to construct the fixtures and surroundings such that this wouldn't be the case).

#### sparky 1

Joined Nov 3, 2018
290
In order to answer which "Shocks the Most" Because it is a combination of both voltage and current both quantities are involved. The question could be reworded. What quantities of voltage and current produce the most shock ?
Understanding the functional relationship between Voltage and Current involves using both terms together in describing an electrical behavior.

What are some of the differences between voltage and current ?
https://www.diffen.com/difference/Current_vs_Voltage#:~:text=Voltage,-Diffen › Science › Physics&text=Current is the rate at,electric current between two points.

(Voltage and Current)
Teaching approach revisited again :

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• howartthou

#### crutschow

Joined Mar 14, 2008
25,247
You need to review Ohm's law.

• Martin_R

#### gerty

Joined Aug 30, 2007
1,291
Speaking of ohms.. Get a few of your friends together, put your meter on MΩ and put one probe in each hand. Squeeze the probe tips and check out the meter reading, now have all your friends do this. You will see the numbers vary a whole lot, and that will have an impact on "what would fry you most"

#### MrSoftware

Joined Oct 29, 2013
1,860
Here's what you're missing. Assuming you have a power supply that lets you set the voltage limit and current limit. You grab the wires and turn the voltage limit to 100V and current limit to 2A, and lets assume the stars align and you're getting 2A at 100V through your body. So to your question, what happens if you turn the voltage down and keep the current the same? This is not possible. As soon as you turn the voltage down, the current will also drop. So you say fine, lets keep the voltage the same and turn the current down. The same applies; if you turn the current down, the voltage will also drop. They are co-dependent. So your theoretical situation is not possible in the real world.

As to what shocks the most, or will kill you; frequency has a lot to do with it. This will be educational:

#### MrChips

Joined Oct 2, 2009
21,634
The answer you are seeking is voltage.

Voltage is e.m.f. = electromotive force. It is the driving force that drives current.
The more voltage you have the larger the current. So think about it. It is both voltage and current. What your body senses is current.

• howartthou

#### MisterBill2

Joined Jan 23, 2018
6,694
The answer you are seeking is voltage.

Voltage is e.m.f. = electromotive force. It is the driving force that drives current.
The more voltage you have the larger the current. So think about it. It is both voltage and current. What your body senses is current.

Like I responded in a much earlier post: It is the voltage that drives the current, and the current is what does the damage.
The comments about constant power situations are an attempt to put it into a non-real universe. UNless the TS is working on the design of a different electric chair for executions, which is probably a forbidden topic on this site. At least I hope it is.
PROBABLY THIS THREAD SHOULD BE DELETED!!!

#### atferrari

Joined Jan 6, 2004
4,069
You need to review Ohm's law.
Yes; all the OP needs.

BTW, George Simon implicit advice is: always watch the voltage.

#### howartthou

Joined Apr 18, 2009
111
Hi All, thanks again. I really appreciate all the responses. Some of them are excellent, especially the video on current vs voltage. This is the essence of my question.

I have to blame myself for not expressing the question clearly enough though, so please let me try again.

I am trying to understand in the power equation P = VI what P really means since we talk about power in watts and use it as a measure of how powerful something is, for example, a 100Kw car engine is more powerful than a 50 car kw engine.

So here is where I get confused. Two circuits with the same power yet one has higher amperage, so shouldn't amps be the measure for power?

I will express the 2 example circuits again using ohm's law as suggested.

First using the power equation P=VI the 2 example circuits are:

10 watts = 10 volts x 1 amp

10 watts = 5 volts x 2 amps

These same circuits can be expressed using ohms law V=IR, respectively:

10 volts = 1 amp x 10 ohms

5 volts = 2 amps x 2.5 ohms

Note that both these circuits have the same wattage (aka power) but one has half the voltage but twice the amps because it has less resistance.

If I had an amp meter on each circuit one would show 1 amp and the other 2 amps. Makes sense so far (I hope).

Now I was trying to find out that since they have the same power but the current (measured as 1 amp or 2 amps) kind of says well the 2 amp circuit is really more powerful because it conducts twice the electrons (controlled by the resistance in the load).

So if the circuits were car engines I would say they have equal power because they are both 10 watts. But how can this be true if one circuit, the 5 volt circuit, is conducting twice as many electrons as the 10 volt circuit. Shouldn't amps be the measure of power because the wattage does not really tell us how much electricity is in the circuit.

I think I know the answer now. It is all about resistance. It just seems silly that 2 circuits can have the same power yet one passes more electricity. Surely the circuit that has twice the electricity is more powerful? Yet amps is not the measure for power. Seems odd to me. It seems that the real power is dependent on the resistance and wattage does not tell you anything about the load so why is it a measure of power?

#### dl324

Joined Mar 30, 2015
11,232
Yet amps is not the measure for power.

$$P = IV = \frac{V^2}{R}=I^2R$$

Whether someone will feel a shock depends on current. A person has resistance, so the amount of current depends on voltage.

5V and 10V are not normally high enough to present a shock risk if you're talking about contact with dry skin. If you're talking about wet skin, contact with muscle tissue (e.g. the heart), etc, a lethal current can be induced by a low voltage.

#### howartthou

Joined Apr 18, 2009
111
5V and 10V are not normally high enough to present a shock risk if you're talking about contact with dry skin. If you're talking about wet skin, contact with muscle tissue (e.g. the heart), etc, a lethal current can be induced by a low voltage.
Can you please forget the electric shock analogy. Its just an analogy, and not a very good one because its distracting from the real question.

My new example talks about watts as a measure of power.

The real question: How can 2 circuits with the same power have different amperage's yet be considered to have the same power?

#### atriano92

Joined Aug 13, 2020
1
I think another way to look at it for your example is that Voltage is the relative difference of potential between two points and current is not. We are in the presence of voltage all the time. Being close to a 100V circuit or 1000V difference of potential will not cause a sensation of shock (just think about ESD which can break circuits without you feeling it). Current through the body WILL cause a sensation of shock (at any voltage), but that will only be possible if the skin is low enough impedance and if the voltage is high enough to cause the current flow. Measures of electric shock and its effect on the human body is generally reported to be a function of the current level (as low as ~0.25mA for sensation, or even above 30mA+ can cause ventricular fibrillation).

The example of watts is simply not applicable when it comes to the sensation of shock. The point of watts is a measure of power and over time, energy. So in your example if the load is an actual resistor (ohms) then both circuits will produce the exact same amount of heat per second and will "burn" you the exact same amount if you touch the resistor.

• howartthou

#### WBahn

Joined Mar 31, 2012
26,047
Hi All, thanks again. I really appreciate all the responses. Some of them are excellent, especially the video on current vs voltage. This is the essence of my question.

I have to blame myself for not expressing the question clearly enough though, so please let me try again.

I am trying to understand in the power equation P = VI what P really means since we talk about power in watts and use it as a measure of how powerful something is, for example, a 100Kw car engine is more powerful than a 50 car kw engine.

So here is where I get confused. Two circuits with the same power yet one has higher amperage, so shouldn't amps be the measure for power?

I will express the 2 example circuits again using ohm's law as suggested.

First using the power equation P=VI the 2 example circuits are:

10 watts = 10 volts x 1 amp

10 watts = 5 volts x 2 amps

These same circuits can be expressed using ohms law V=IR, respectively:

10 volts = 1 amp x 10 ohms

5 volts = 2 amps x 2.5 ohms

Note that both these circuits have the same wattage (aka power) but one has half the voltage but twice the amps because it has less resistance.

If I had an amp meter on each circuit one would show 1 amp and the other 2 amps. Makes sense so far (I hope).

Now I was trying to find out that since they have the same power but the current (measured as 1 amp or 2 amps) kind of says well the 2 amp circuit is really more powerful because it conducts twice the electrons (controlled by the resistance in the load).

So if the circuits were car engines I would say they have equal power because they are both 10 watts. But how can this be true if one circuit, the 5 volt circuit, is conducting twice as many electrons as the 10 volt circuit. Shouldn't amps be the measure of power because the wattage does not really tell us how much electricity is in the circuit.

I think I know the answer now. It is all about resistance. It just seems silly that 2 circuits can have the same power yet one passes more electricity. Surely the circuit that has twice the electricity is more powerful? Yet amps is not the measure for power. Seems odd to me. It seems that the real power is dependent on the resistance and wattage does not tell you anything about the load so why is it a measure of power?
To use your car engine analogy, power is the product of the torque on the drive shaft and the rotational speed of the drive shaft. What you are basically trying to ask is which is more important, torque or speed. It's not that simple and it's the product that counts (for determining the power). I can have a very small engine turning real fast (think something like a chainsaw engine) but that produces very little torque and hence very little power, or I can have something that turns very slowly and produces a lot of torque (think something like the steam engines on the Titanic) and hence lots of power. But I can easily go the other direction and have something that spins very fast with a modest amount of torque (like the motor in an Indy race car) outputting much more power that something that outputs more torque but at a much lower speed (perhaps the motor on a car hoist).

In the case of electrical current, it is NOT the case that more current means more power. The voltage determines how much energy EACH electron delivers, so having 2 A of current in which each coulomb of charge delivers 10 J of energy will deliver considerably less power than having 10 A of current but where each coulomb delivers only 0.1 J of energy.

But these discussions are very separate from the question of what is more "dangerous" in terms of shock potential. The dangers associated with a shock are due to the current that passes through sensitive parts of the body disrupting the body's normal processes, such as the electrical impulses that tell the heart to beat (or exactly how to beat). The interference doesn't have to have much to do with the brute electrical power that is delivered.

• howartthou