How does high voltages of electricity coming from power plants also helps in saving energy?
Electronic – High voltages vs. Energy loss
energyvoltage
Related Solutions
If you are only confused by the "test particle," then you can think of it similarly to a multimeter. With a multimeter, you can probe a circuit to determine a voltage at one part of a circuit relative to another. With a test particle (or test charge, as I got used to hearing), you place it at a point in space, and "observe" it's behavior to see how electric (or magnetic) fields are oriented.
Like charges repel each other, so if a test charge would tend to move in a certain direction in space, then either that direction contains a negative charge (assuming you use a positive test charge), or the opposing direction contains a positive charge.
The movement of a test charge will always oppose the energy gradient (in three dimensions, energy is a scalar field, so the spatial derivatives are your forces, since energy divided by length is force). Thus, a test charge will move in the direction that achieves it's lowest energy state.
In a vacuum (such as space), ionized particles can move freely. The test charge is usually assumed to be in a state such that it can move freely. This doesn't necessarily correlate to anything real, but is a hypothetical state so the fields can be analyzed easily. You are correct in that circuits don't involve the movement of atoms or positive charges, but rather electrons (due to d-block delocalization, but that's chemistry) move. The positive charges (protons) are held in a crystalline lattice, which is why they don't move. In a conductor, electrons can move freely, so they move in response to an applied electric (or changing magnetic) field.
In free space, however, a test charge (which is really an ion, or a proton, or a positron, or a myriad of other positively charged particles) is not constrained by the bonds that hold metal atoms in place, so it can move in response to an applied field. Specifically, interactions governed by photons cause particles to exchange energy, creating the field gradient mentioned before. Therefore, test charges in free space move similarly to how electrons move in a metal (or another conductor).
While this is a convention (positive charges just seemed to be more reasonable when a lot of this math was derived), you could physically create a positively charged particle and observe it's trajectory. You could even do this at home. You'd be using a Cloud Chamber and a Beta emitter to see them.
I hope this was helpful! Let me know if you need any more clarification.
Having designed a number of energy meters over the years, I can say with authority that reactive energy (Q) is not the same as exported energy (PE), in fact it has basically nothing to do with it. What you need is an energy meter that has separate counters for imported real energy (P) and exported real energy (PE).
Reactive power is basically only relevant for industrial metering, but in case you're interested, an energy meter can additionally have up to four counters for reactive energy (Q), these being, in order of real world relevance, inductive (QIND) and capacitive (QCAP), exported inductive (QEIND) and exported capacitive (QECAP). When there are six registers in total, the sign of real power is used to select which registers are accumulated.
Most important, however, is that when money is involved (the meter is used for billing), the exact meter brand and model must be accepted by your power grid company. If money is involved, they may also come and seal the meter themselves with their own seal, although the practice varies from company to company in different countries.
When you're reading the registers, be sure to note the difference between instantaneous power (P) and accumulated energy (W). The register names for instantaneour powers might be something like P, PE, Q, QIND, QCAP and the energies might be something like W, WE, WQ, WQIND, WQCAP etc.
Best Answer
The power cables have some resistance. Power lost in the wires can be calculated as
P=R*I^2
with R as the resistance of the wires and I as the current that passes through them. Power at the load isP=V*I
. From this you can see that if you increase the voltage by 2x, you need only half the current to deliver the same power. However, if you pass half the current on the same wires, you will lose only a quarter of the power.An example with made up exaggerated numbers:
Wire resistance is 10 Ohm (not very long power line), power required is 100kW. At 220V, the required current will be
100kW/220V=454A
. The power lost at the wire will be10Ohm*454A^2=2.061MW
. Voltage lost at the wire will be10Ohm*454A=4540V
. So, the power plant will have to generate 4760V so that the load can receive 220V. Power generated at the plant - 2.061MW+0.1MW=2.161MW. Efficiency is0.1MW/2.161MW*100%=4.6%
Now suppose the plant generates 330kV (and the voltage is stepped down to 220V very close to the load). Now the current needed is only
100kW/330kV=0.3A
. Power lost on wires is10Ohm*0.3A^2=0.9W
, so the efficiency (ignoring power lost in the transformer) is100000W/100000.9W*100%=99.9991%
. Transformers are very efficient, about 98% for the big ones used in power distribution.Now scale that up to, say, a power plant that generates 1GW and you will see why Tesla won the current wars (hint: there were no efficient DC-DC converters back then).