Why does the zero line get charged, so that the neon tube of the electroscope glows
It involves the following knowledge points:
1. The principle of neon tube lighting
The neon tube glows not because the circuit under test (the charged body under test) has current flowing, but because the charged body under test has voltage.
In fact, there is a parallel relationship between the electric pen and the charged body under test.
When the measured point has a voltage to the ground (greater than 0v), the voltage flows back to the ground through the tip of the electric pen, the safety resistor, and the human body, forming a path, and the current flowing through the neon tube makes the neon tube glow.
The neon tube can only emit light when a large enough current flows. If the potential difference (voltage) of the measured point to ground is not 0, but it is not too large, the neon tube cannot be made to emit light. Of course, the current that makes the neon tube light is Very small, for the human body, is a safe current.
Therefore, when the electric pen emits light, what the electric pen tests is actually voltage, not current. It has nothing to do with whether there is current flowing through the measured point.
That is to say, there is a voltage to ground to make the electric pen drive light.
2. When the circuit is working normally, why does the electric pen not light up when it touches the zero line?
It seems obvious, but it's not.
When the circuit is working normally, since the neutral line is connected to the "neutral point" of the transformer, the voltage (potential) of the neutral line to the ground is 0, and the electric pen tests the voltage, so the electric pen does not emit light.
But at this time, the current on the neutral wire is exactly the same as the current on the live wire!! It is even very large!! That is, it is not that there is no electricity on the neutral wire, and the neutral wire has no electricity, but the voltage of the neutral wire to ground. due to the theoretical 0
3. Why is the voltage of the neutral line 0?
The zero line voltage is 0, which is not accurate in practice, but is theoretically 0, which is always greater than 0v in practice.
In the home power supply circuit, it is usually two-phase (single-phase electricity), which are connected to the power supply transformer through the transmission line, and the power supply transformer is three-phase alternating current. "Neutral", then all the neutrals of the power supply lines are connected from the neutral.
Because of the "neutral point" theory, when the three-phase circuit is balanced, that is, when the loads of the three-phase circuits are exactly the same, the ground potential of this neutral point is 0. This is only theoretical, in fact, the three-phase electrical load It cannot be absolutely equal. Therefore, the ground potential of the "neutral point" is always greater than 0, which means that the ground potential on the neutral line of all household lighting transmission lines is always greater than 0.
Then why is the zero line voltage not 0, and the neon tube of the electroscope does not emit light? It is because the (to ground) voltage on this zero line is very small in normal circuits, and it passes through the safety resistance of the electroscope and After the resistance of the human body, the current is extremely weak, and it is not enough to drive the neon tube to emit light.
4. Under what circumstances is the voltage of the neutral line not 0, and it is larger?
There are usually three cases,
First, when the neutral point of the transformer is not grounded, the neutral point is not grounded (floating), (although the current of the neutral point is 0, but if it is not connected to the ground, the potential between the neutral point and the ground Still will not be automatically equal.), so that the ground potential of the neutral point is not 0, at this time, the middle tap of the three-phase power has no meaning of "neutral point". Make all lighting power supply circuits output by the transformer zero When the neutral line fails, even if the appliance has a protective grounding, it will be invalid, because after the current including the grounding flows to the ground, the neutral point of the transformer is suspended, so that the protective current cannot flow back to the ground loop. So people will still get an electric shock after touching them.
The second is that the unbalance of the three-phase load is too large, causing the neutral point to "drift". At this time, the potential difference between the neutral point and the ground is not 0, and it is relatively large. At this time, even for a normal working circuit , the voltage on the neutral wire will also make the neon tube of the electric pen glow.
The third is the circuit failure of the lighting circuit itself, the most common is the short neutral line behind the lamp holder (electrical appliance, etc.). Because of the short neutral wire, for example, the side connected to the lamp holder is point A, and the side connected to the transformer circuit It is point B. Then point B is directly connected to the ground, and its potential difference to the ground is 0, so the electric pen does not emit light. But at this time, since point A is not connected to the ground, its voltage is not 0, and it is relatively large, so, it is At this voltage, the neon tube that drives the electric pen emits light. However, this light emission is still less bright than the neon tube of the phase line, because after all, the voltage at point A is still much smaller than 220V, but it can drive the neon tube to emit light.
5. How does the voltage at point A on the neutral line come from?
It's electromagnetic induction!
Induction, phenomena are everywhere! Human body induction, telepathy...
Since there is an alternating voltage and an alternating magnetic field on the live wire, an induced voltage will be generated on the neutral wire and on the neutral wire of the breakpoint of the A terminal.
From the perspective of stray capacitance and parasitic capacitance, the alternating voltage on the live wire will also generate an induced charge on the neutral wire of the breakpoint. Thereby an induced voltage is generated....
