# Voltage gain vs power gain

Gold Member
Hey guys, I need confirmation on this.
Since V=IR and P=IV....and V^2/R=P
Lets say your body has a resistance of 2000 ohms and you get shocked by 120 volts. You receive 7.2 watts of power.

Now lets step to the left and get shocked by 480 volts with the same 2000 ohm body.
I now get 115.2 watts of power or roughly 16 times the electrocution of 120 volts.
Would you agree? Obviously, the assumption is that you would only feel 4 times the electrocution which I believe is wrong.

Same thing for arc flash. Bang two 120 volt wires together and you see a ball of flame about the size of your fist. (obviously can be deadly, do not try)

Bang two 480 volt wires together and you see 16 times the arc flash instead of the 4 times the arc flash you might think. (obviously limited by its transformer if applicable) (arc flash extremely deadly at 480, you will die, do not try)

In the arc flash example I am also assuming the same resistance of the copper and buss and so forth. According to V^2/R=P......my theories are correct....What do you think?

So 2400 volt would give 400 times the electrocution and arc flash as say 120 volt.
I can't help but notice this pattern (20)^2=400. Voltage gain squared equals power gain when using the same resistance.

## Answers and Replies

Power is proportional to the square of the voltage.

I don't know that electric sensations work on a linear scale.

I suspect (but don't know) that arc flash also scales with the square of the power. I'm basing this on the heat dissipation going up with the area of the plasma ball, but it could also go up with color (light frequency) which is a 4th power effect. So under my assumption arcflash would scale with voltage.

Hesch
Gold Member
An arc is not a resistance, but rather a zenerdiode.

At short distance between two electrodes ( say 1 mm ), when you increase the voltage across the airgap, the field strength may come up to 2 kV/mm before the arc will flash. Having established this arc, the voltage across the arc will be about 20 V.

Now say that the supplied voltage is 2000 V and the impedance of some voltage source is 50Ω, the current through the arc will be:

I = ( 2000 V - 20 V ) / 50Ω.

I don't think that the characteristic of a human body can be compared to a resistor or a zenerdiode.

anorlunda
Staff Emeritus
Lethality is not linear, no matter what your unit of measure, voltage, current, power. The linked article has some interesting info.

https://en.m.wikipedia.org/wiki/Electric_shock#Pathophysiology

Log-log graph of the effect of alternating current I of duration T passing from left hand to feet as defined in IEC publication 60479-1.[18]
AC-1: imperceptible
AC-2: perceptible but no muscle reaction
AC-3: muscle contraction with reversible effects
AC-4: possible irreversible effects
AC-4.1: up to 5% probability of ventricular fibrillation
AC-4.2: 5-50% probability of fibrillation
AC-4.3: over 50% probability of fibrillation​

sysprog
Gold Member
Interesting stuff.

Can we all agree with this statement?
"Voltage gain squared equals power gain when using the same resistance"

In other words, lets say we have a 6 ohm heater and I hook it to 120 volts. We get 20 amps at 2400 watts.

Now lets take that same 6 ohm heater and I hook it to 240 volts. We now get 40 amps at 8400 watts. (2^2)=4 times power gain when compared to 120v.

Now lets takes the same 6 ohm heater and I hook it to 480 volts. We now get 80 amps at 38,400 watts. (4^2)=16 times power gain when compared to 120V.

With same 6 ohm resistor, 2400 volts will then give (20^2)= 400 times the power of 120 volt. Etc.

Can we all agree with this? Anyone disagree? Obviously I am assuming a very robust resistor that can handle any voltage that internal resistance does not change.

Gold Member
Also, people argue if its the volts or amps that kill you in electrocution as far as burning. Sure its a combination of the two, but I believe its the WATTS that kill you. Isn't WATTS a measure meant of work being done? Isn't 100 watts four times the work as 25 watts? Isn't 4 HP four times the power as 1 HP? What's the difference?

anorlunda
Staff Emeritus
Also, people argue if its the volts or amps that kill you in electrocution as far as burning. Sure its a combination of the two, but I believe its the WATTS that kill you. Isn't WATTS a measure meant of work being done? Isn't 100 watts four times the work as 25 watts? Isn't 4 HP four times the power as 1 HP? What's the difference?

Did you read the wikipedia article? It's not that simple.

Just saying that you believe something without giving the foundation won't get you very far on PF.

By the way, work is energy, Watt-hours are energy, and watts are the rate of delivery of energy. That's where the time axis on the lethality chart in #4 comes in.

jim hardy
Science Advisor
Gold Member
Dearly Missed
Keep it simple

P = E2 / R
if you make R constant
p is in proportion to square of voltage

but as has been mentioned,
an arc is a LONG WAY from constant resistance.
Ever electric welded?? Then you've felt it.

Draw from your everyday experience.
eric hoffer said:
"I have never been able to think things out. I have to live them out, thinking as i go along."

Gold Member
Keep it simple

P = E2 / R
if you make R constant
p is in proportion to square of voltage

but as has been mentioned,
an arc is a LONG WAY from constant resistance.
Ever electric welded?? Then you've felt it.

Draw from your everyday experience.

Thanks guys, I read the wikepedia article and I get the jist of what you are saying overall. I can see the graph above is on logarithmic paper rather than lineal.
My point in general is that if you are standing next to a wide open energized 480 panel (with the face plate removed).....you should be nearly "soiling" yourself instead of being relaxed and comfortable, especially as an engineer. 480 ARC flashes are huge and powerful!

To Jim Hardy. Jim, I have done just about everything in life except two things. Played hockey and arc welded.
Shame on me.

jim hardy
Science Advisor
Gold Member
Dearly Missed
To Jim Hardy. Jim, I have done just about everything in life except two things. Played hockey and arc welded.

I never played hockey either.. but then i did grow up in Miami , unaware water could freeze outdoors.

No shame in never having needed to electric weld. I was introduced in 9th grade shop class.

I recommend the classic homeowner Lincoln cathedral buzzbox. Arc stays lit way better than those frustrating 110 volt cheapies.
Around \$100 at yardsales

Once you've owned one you'll never again be able to live without.

i made an extension cord to use the dryer outlet.

old jim

anorlunda
Staff Emeritus
Here's a more nuanced view. It plots time versus current. But if current and volts are proportional, the horizontal axis could be power in watts. Time*power = energy measured in joules. The upper right corner in that plot is the most energy.

https://en.wikipedia.org/wiki/Electrical_injury

DaveE
Science Advisor
Gold Member
This is why the Europeans put a LOT of fuses in their 240 V plugs.
No! Fuses are neither fast enough or accurate enough to protect against electric shock. In Europe, US, or anywhere else, they are primarily used to prevent fire. In modern equipment, European designs don't have more (or less) fusing. They are also used to protect equipment from heat damage due to fault currents, like stalled motors. In distribution networks, they can be designed to isolate overloaded portions of the network to prevent large scale failures.

Protection against electric shock is primarily from grounded enclosures or reliable insulation systems. Also fast electronic systems like GFCI.

Also, your answer is about 5 years too late.

[Mentor Note -- The quote in this post is from a reply that has since been deleted for misinformation. Good reply by @DaveE ]

Last edited by a moderator:
rbelli1
Gold Member
Also fast electronic systems like GFCI.
And arc fault detection. I was involved in an arc flash situation and was saved from serious injury or death by some fancy Japanese safety equipment. 1000A 480V three phase disconnected in a fraction of a second after a metal powder short.

BoB

berkeman
Power gain is similar to voltage gain; it is the ratio of output power to input power and can also be represented in the log domain.

For example, if input power is 0 dB and output power is 10 dB, the dB domain’s power gain is 10 dB.

Voltage gain is the ratio of output voltage to input voltage, measured in log domain as shown: In voltage gain a factor of 20 is multiplied to log, this is because Pout=Vout2/R, the 2 in log domain gets multiplied to 10 to get 20.

sophiecentaur
Science Advisor
Gold Member
2020 Award
Power gain is similar to voltage gain; it is the ratio of output power to input power and can also be represented in the log domain.

View attachment 288847

For example, if input power is 0 dB and output power is 10 dB, the dB domain’s power gain is 10 dB.

Voltage gain is the ratio of output voltage to input voltage, measured in log domain as shown: In voltage gain a factor of 20 is multiplied to log, this is because Pout=Vout2/R, the 2 in log domain gets multiplied to 10 to get 20.

View attachment 288848
But don't forget that the R must be the same throughout. It very often isn't - take a transformer for instance. Free gain? Really?