Suppose you have a 120 volt, 20 amp circuit which only has a light bulb connected. When measuring the voltage going to the light bulb my meter reads 120 volts. The basics taught me that a circuit is a circle, so the back and forth movement of electrons continues after the light bulb along the neutral wire back to the socket etc. So, why isn't there voltage on the neutral wire? I'm assuming there has to be current in order for the circuit to work. If so, is there current without voltage? Thanks for any help with this!
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5$\begingroup$ If you're asking this question, you shouldn't be measuring or working with 120VAC circuits at all. Please stop. $\endgroup$– Alfred CentauriNov 20, 2013 at 23:11
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$\begingroup$ Related: physics.stackexchange.com/q/80400/2451 and links therein. $\endgroup$– Qmechanic ♦Nov 20, 2013 at 23:49
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1$\begingroup$ Alfred, is this how you answer all your student's questions when you occasionally teach your EE classes? $\endgroup$– user1332449Nov 21, 2013 at 0:25
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$\begingroup$ @user1332449, your question doesn't logically follow from my comment. Do you suppose that I would tell a student not to work with 120VAC if she asks how to find the Thevenin equivalent circuit or how to do voltage division or how to find the small signal output resistance of a BJT amplifier? $\endgroup$– Alfred CentauriNov 21, 2013 at 1:12
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3 Answers
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Voltage is the difference in potential between two wires. You can't say the "hot" wire has any sort of voltage by itself. Just as you can't say the "neutral" wire has any voltage by itself. The voltage is the measure of electrical potential between the two.
When you connect the light bulb between the two it provides a path for current to flow from high voltage to low voltage.
If you want to read about the specifics of this, Kirchoff's circuit laws (and especially Kirchoff's Voltage Law) are the place to start.
answered Nov 20, 2013 at 22:56
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$\begingroup$ thanks, so when I measure the voltage of the hot I see 120v relative to ground or neutral. All that voltage hits the lightbulb correct? Then, the return from the lightbulb to the socket on the neutral has no difference in potential and therefore measures the same relative to ground--zero volts. Yet electrons are still moving...correct? $\endgroup$ Nov 20, 2013 at 23:58
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$\begingroup$ The full amount of voltage between the hot and neutral is across the light bulb. If you measure that voltage as 120 V then the voltage drop across the light bulb will be 120 V. I have edited the answer with a link. $\endgroup$ Nov 21, 2013 at 0:15
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Actually there is a little voltage on the neutral wire when current is flowing thru it. However, this wire has low resistance specifically so that the voltage accross it remains small. The resistance of the lightbulb is much larger than that of the wire, which is why you see most of the voltage accross the bulb and not the wire.
Consider the limiting case where you have a short instead of a lightbulb. This puts half the voltage accross the neutral wire (the other half on the hot wire because both wires are in series), which causes so much current to flow that something would blow up or melt or catch fire if allowed to persist. That's why there is a fuse or circuit breaker in series with the hot wire. The fuse trips or the circuit breaker pops when the current gets above normal levels, but before this high current lasts long enough to cause your house to catch fire.
For example, let's say that both the hot wire and the neutral wire each have 1 Ω resistance back to the bus bar in the breaker panel. If the voltage is 110 V and you use that to power a 60 W lightbulb, about 550 mA will flow. That will cause a voltage drop of 550 mV on each of the hot and neutral wires. The bulb will actually see 1.1 V less because of this, but that's a small fraction of 110 V and less than the normal variation of the power line, so you won't notice it. In this example, if you measured the voltage between the neutral at the light bulb and the third ground wire, you would see the 550 mV.
answered Nov 20, 2013 at 23:45
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$\begingroup$ Thanks Olin, Looking solely at the current in the circuit, if your exampled 550mA are flowing to the light bulb, then how do you calculate what is flowing on the return? Does Kirchoff apply here resulting in 550mA? $\endgroup$ Nov 21, 2013 at 0:40
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1$\begingroup$ Actually there is a little voltage on the neutral wire when current is flowing thru it. there is little voltage relative to ground but there is 120VAC on it relative to hot. We need to carefully distinguish between the voltage on (relative to something else) the conductor and the voltage across the conductor. This is, I think, the root of the OP's confusion. $\endgroup$ Nov 21, 2013 at 1:46
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$\begingroup$ Alfred, thanks for your clarification--I understand what you are saying. So, to wrap all of this up, do you know how much current is on that neutral leg when current is flowing through it (when the light is on)? $\endgroup$ Nov 21, 2013 at 16:48
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$\begingroup$ @user: The hot wire, load, and neutral wire are all in series, so the same current flows thru all of them. In the case of the lightbulb example, that would be 550 mA thru each of them. $\endgroup$ Nov 21, 2013 at 18:22
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$\begingroup$ So, in summary, we're looking at the same amount of current with a much smaller amount of voltage across the neutral conductor. Please correct me if I'm wrong. Thanks for everyone's help. $\endgroup$ Nov 21, 2013 at 18:42
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Wires are in fact resistors, but with VERY VERY tiny amounts of energy being thermally dissipated by the current due to EXTREMELY low drops in voltage over large lenths of the wire. Thus, current DOES flow through the neutral wire, but the drop in potential along a length is literally far too small for your voltmeter to detect. Review Kirchoff's Voltage Law and you will see the rise in potential in a closed loop circuit is equal to the drop in potential for all elements along the closed loop; NOTE - voltage across the battery is equal but, opposite in polarity to that across the resistors (or in your case a light bulb which IS a resister with a high rate of thermal dissipation which is why it lights up!). However, in reality, a very tiny drop in potential DOES occur in the wire, but is so small it is effectively ignored as it very rarely has any pronounced affect on your calculations or circuit design unless you have chosen a very inappropriate wire for your circuit. If you chose a wire as big around as your arm, then the resistance would likely be SO high that no current would flow and your bulb would not light up. The entire drop in voltage would occur as a combined serial resistance of your light bulb and the wire itself.
