# Thread: Why I get shocked touching 1 wire of circuit and not getting shocked touching the other

1. imgur: the simple image sharer If electricity doesn't travel to ground why I get electrocuted when touching wire closer to the "+" symbol and nothing happens when I touch the wire closer to the "-" symbol? How are these wires different and why? Is wire closer to "-" called the hot wire and wire closer "+" neutral or otherwise?

2.

3. Originally Posted by fghf76
imgur: the simple image sharer If electricity doesn't travel to ground why I get electrocuted when touching wire closer to the "+" symbol and nothing happens when I touch the wire closer to the "-" symbol? How are these wires different and why? Is wire closer to "-" called the hot wire and wire closer "+" neutral or otherwise?
In the circuit shown, the circuit is complete without any current flowing through the ground. It just flows from the battery positive, through the load and back to battery negative. If the ground symbol on the drawing represents an earth ground, and if you are standing on the earth, or touching a water pipe or something else connected to the earth, now there is a possible path from the battery positive, through your body, to the earth, then back to the battery negative.

On the other hand, if you touch the negative wire, no current flows through your body because there isn't any potential difference between the negative side of the circuit and earth.

In a d-c circuit, the word "neutral" is not used. That is a term used for the grounded side of a transformer in an a-c circuit. In the US, this is the center tap of the secondary side of a 480/120-240 volt transformer.

4. Why is voltage different on wire closer to negative and wire closer positive (I can't imagine why is voltage different when coming out of lightbulb if it is the single circuit)

5. Ohm's law. Current through a resistor results in a drop in voltage.

6. But if electrons travel from - to + shouldn't voltage be less on a positive wire instead of higher?

7. Originally Posted by fghf76
imgur: the simple image sharer If electricity doesn't travel to ground why I get electrocuted when touching wire closer to the "+" symbol
You won't get shocked by either wire at 1 volt. If that was 1000 volts and you were not grounded you wouldn't get a shock from either one. But if you are grounded, electrons will flow from ground through your finger into the + lead, and you would get a shock.

and nothing happens when I touch the wire closer to the "-" symbol? How are these wires different and why? Is wire closer to "-" called the hot wire and wire closer "+" neutral or otherwise?
"Neutral" is a specific wiring term and doesn't mean much in terms of potential. Usually, but not always, it is close to ground.

8. Originally Posted by fghf76
But if electrons travel from - to + shouldn't voltage be less on a positive wire instead of higher?
What the voltage is at any particular point of the circuit is really an arbitrary choice. We tend to assign the value of zero voltage to the ground point for convenience. But what really counts is just the voltage difference between two points. So if you are standing in contact with the ground and the negative side of the battery is grounded, then you have a larger voltage difference across your body when you touch the positive wire than if you touch the negative. What you have done is give the electrons a new path from - to + to follow.

9. Originally Posted by fghf76
But if electrons travel from - to + shouldn't voltage be less on a positive wire instead of higher?
Yes, the electrons flow from negative to positive. That's just a convention that was chosen many years ago. But, the electrons will shock you just as much no matter which way they flow through your body.

10. In the circuit you are showing you have a 100ohms resistance (lamp).
If you are contacting ground on the negative side of the lamp which is also grounded you are at virtually the same voltage as the ground is.
When you are contacting ground and contact the circuit on the positive side of the load you have the full difference between the ground and whatever voltage made it past the 100ohms lamp.
In effect you are turning yourself into a short circuit across the terminals of the load.
You will conduct a current (get a shock) through you that is inversely proportional between your body's resistance and the 100ohm resistance.
It is the amperage that kills, not the voltage.

By the way, a dry skin gives you about 10,000ohms but wet skin drops that to about 1,000ohms.
You can feel about 1 milliamp of ac current and about 4 milliamp of dc current. Over 10 milliamp your muscles start to contract and if it is across the heart it can be lethal.

11. Originally Posted by dan hunter
In the circuit you are showing you have a 100ohms resistance (lamp).
If you are contacting ground on the negative side of the lamp which is also grounded you are at virtually the same voltage as the ground is.
When you are contacting ground and contact the circuit on the positive side of the load you have the full difference between the ground and whatever voltage made it past the 100ohms lamp.
If he's contacting the positive side, the 100 ohm lamp does not come into play. He's getting the full battery voltage.
In effect you are turning yourself into a short circuit across the terminals of the load.
You will conduct a current (get a shock) through you that is inversely proportional between your body's resistance and the 100ohm resistance.
It is the amperage that kills, not the voltage.
The current will be proportional to the voltage and inversely proportional to the body's resistance. The 100 ohms doesn't matter.
By the way, a dry skin gives you about 10,000ohms but wet skin drops that to about 1,000ohms.
You can feel about 1 milliamp of ac current and about 4 milliamp of dc current. Over 10 milliamp your muscles start to contract and if it is across the heart it can be lethal.

12. Originally Posted by fghf76
But if electrons travel from - to + shouldn't voltage be less on a positive wire instead of higher?
Listen to Harold.. he is right.. and might I sound a clear warning.. DO NOT be messing about with electrical currents..
YOU can get very hurt.. big time, dead even.. Direct current and Alternating circuitry are different.. Learn, and research.
~ Ask but do not touch that what you do not understand.. I saw you ask of why the voltage drops.. some of it was converted to heat or motion.. On a A/C system both phase and Neutral are live.. On a D/C circuit only one wire is loaded.. It can be a very complex subject..
Seek advice of professionals.. Looking at your image.. There is a confusion of why it would be grounded at all.? It's a closes circuit.

13. Originally Posted by Harold14370
Originally Posted by dan hunter
In the circuit you are showing you have a 100ohms resistance (lamp).
If you are contacting ground on the negative side of the lamp which is also grounded you are at virtually the same voltage as the ground is.
When you are contacting ground and contact the circuit on the positive side of the load you have the full difference between the ground and whatever voltage made it past the 100ohms lamp.
If he's contacting the positive side, the 100 ohm lamp does not come into play. He's getting the full battery voltage.

In effect you are turning yourself into a short circuit across the terminals of the load.
You will conduct a current (get a shock) through you that is inversely proportional between your body's resistance and the 100ohm resistance.
It is the amperage that kills, not the voltage.
The current will be proportional to the voltage and inversely proportional to the body's resistance. The 100 ohms doesn't matter.
By the way, a dry skin gives you about 10,000ohms but wet skin drops that to about 1,000ohms.
You can feel about 1 milliamp of ac current and about 4 milliamp of dc current. Over 10 milliamp your muscles start to contract and if it is across the heart it can be lethal.
Harold, if he is only touching one wire and is isolated from anything else he would feel no shock (except maybe for a slight static spark because of his capacitance) so for him to feel a shock he would have to be touching two points in the circuit with different voltages.

Take another look at the diagram he posted the link to. They give no resistance for the wire, they grounded the negative terminal and inserted a light with a 100 ohm resistance.
Without the 100 ohm resistance you wouldnt get a shock even if you touched the battery terminals because there would not be any difference in the voltage at those points.

14. I will just add that I have many years of electrical work on board ships.. where everything is a closed system.. No on board earth is used..
My concerns for your safety might be overstated if you are only messing about with little voltages and batteries ie; 1.5 to 9 volts..
I only assumed you were talking of household voltages.. pleased to be wrong..

15. Originally Posted by dan hunter
Originally Posted by Harold14370
Originally Posted by dan hunter
In the circuit you are showing you have a 100ohms resistance (lamp).
If you are contacting ground on the negative side of the lamp which is also grounded you are at virtually the same voltage as the ground is.
When you are contacting ground and contact the circuit on the positive side of the load you have the full difference between the ground and whatever voltage made it past the 100ohms lamp.
If he's contacting the positive side, the 100 ohm lamp does not come into play. He's getting the full battery voltage.

In effect you are turning yourself into a short circuit across the terminals of the load.
You will conduct a current (get a shock) through you that is inversely proportional between your body's resistance and the 100ohm resistance.
It is the amperage that kills, not the voltage.
The current will be proportional to the voltage and inversely proportional to the body's resistance. The 100 ohms doesn't matter.
By the way, a dry skin gives you about 10,000ohms but wet skin drops that to about 1,000ohms.
You can feel about 1 milliamp of ac current and about 4 milliamp of dc current. Over 10 milliamp your muscles start to contract and if it is across the heart it can be lethal.
Harold, if he is only touching one wire and is isolated from anything else he would feel no shock (except maybe for a slight static spark because of his capacitance) so for him to feel a shock he would have to be touching two points in the circuit with different voltages.

Take another look at the diagram he posted the link to. They give no resistance for the wire, they grounded the negative terminal and inserted a light with a 100 ohm resistance.
Without the 100 ohm resistance you wouldnt get a shock even if you touched the battery terminals because there would not be any difference in the voltage at those points.
That's true. Of course, then you would have a dead short on the battery. I'm just noting that the 100 ohms does not factor into the calculation of how much current goes through the person's body, unless you want to consider the voltage drop across the internal impedance of the battery which is usually negligible.

16. Originally Posted by Harold14370
....Of course, then you would have a dead short on the battery. I'm just noting that the 100 ohms does not factor into the calculation of how much current goes through the person's body, unless you want to consider the voltage drop across the internal impedance of the battery which is usually negligible.
But in this case it is a parallel resistance circuit and Kirchoff's laws state the current leaving the circuit equals the current entering the circuit.....

Ah, I think I got you. You are taking only the branch current instead of the total circuit current.
Yes, the current through each resistor is calculated using ohm's law even though the parallel resistors are splitting the total current.
Yes, you could take it as a dead short across the supply.

17. Originally Posted by fghf76
Why is voltage different on wire closer to negative and wire closer positive (I can't imagine why is voltage different when coming out of lightbulb if it is the single circuit)

The voltage gradually reduces along the wire, because the wire has some resistance, albeit very low.

And V = I x R, (Voltage = amps x resistance). In this case, assuming reasonably sized low resistance wire; nearly all of the 1 Volt supply is dropped across the bulb, so there must be 0.01 Amp flowing round the circuit, found by rearranging the Ohms law equation to give I = V/R.

If you had a very sensitive and accurate voltmeter, you could measure the voltage drop along the wire. At the battery + terminal, it will be 1V. At the + side of the bulb, the voltage will be something like 0.999V. At the - side of the bulb, the voltage will be something like 0.001V and at the battery - terminal, 0V.

Voltage is the pressure of the electricity, like water pressure in a pipe. Current is a measure of the actual flow of electrons. If you measured the water pressure at the tap, it would be more than the pressure of the water flowing out the end of a hosepipe. If you plugged the end of the hosepipe, so no water flowed, the water pressure would then become the same everywhere in the pipe. The flow of water through a resisting pipe, causes the pressure drop. Voltage in a circuit drops in the same way when current flows through resistance.

You won't get a shock from 1 volt, but 50 milliamps through your heart can kill you. If part of your body was touching the same ground as the circuit, and you touched the battery + terminal, current would flow through your body. But it would depend on the resistance of your body as to how much current would flow.

OB

18. Originally Posted by One beer
Originally Posted by fghf76
Why is voltage different on wire closer to negative and wire closer positive (I can't imagine why is voltage different when coming out of lightbulb if it is the single circuit)

The voltage gradually reduces along the wire, because the wire has some resistance, albeit very low.
If you are going to consider the resistance of the wire, you should also mention the internal resistance of the source. The source can be treated as an ideal source in series with a resistance. So the voltage of the ideal source is dropped by the load, wire and internal resistance. Thus, if you were to short across the load, you end up with a voltage divider made up of the wire and internal resistance. You'll see a decrease in the voltage seen across the source, due to the larger voltage drop of the internal resistance.

19. If you are going to consider the resistance of the wire, you should also mention the internal resistance of the source. The source can be treated as an ideal source in series with a resistance. So the voltage of the ideal source is dropped by the load, wire and internal resistance. Thus, if you were to short across the load, you end up with a voltage divider made up of the wire and internal resistance. You'll see a decrease in the voltage seen across the source, due to the larger voltage drop of the internal resistance.[/QUOTE][/QUOTE]

Yup, agreed. However, the OP obviously has only a very basic understanding of electronics, (no disrespect intended to them), so I wanted to keep it relatively simple.

Source impedance and internal resistance are not easy concepts to understand unless one has a good grounding in electronics.

OB

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