Basic Electricity Questions

quiquequique Solar Expert Posts: 252 ✭✭
Here we go again...more theory.

I get the water - electricity analogy, a bit. I just want to make sure of my understanding of it. Ive read you equate water flow (m3/s, for example) to amperage, and water pressure to voltage.

1. This would mean the phrase "you've got 120V coming out of that wall socket" is conceptually incorrect. Voltage would be the potential energy between two points, such as a waterfall having potential energy from its top (at say, 100m) and its bottom (at 0m). And thus voltage is a differential of electric energy basically between a non live conductor or earth and a live conductor carrying a potential of 120V?

It is not until electricity is allowed to flow (waterfall water actually falls from 100 to 0) that amperage (flow rate) is produced.

2. In terms of a solar panel V-I graph such as this:

Attachment not found.

as voltage, that differential, gets bigger, current begins to decrease until it gets so big that no current can flow anymore. Until it reaches an open circuit voltage which is analogous to opening up a circuit where no more current flows.

3. Isc starts off at zero when there is an open circuit, but as you close the circuit and current begins flowing, as the differential gets smaller, more electrons can move until its equivalent to shorting the circuit.

4. Finally the Vmp and Imp is the optimum point at which V & I are optimum.

5. On a related note, if Im on DC circuit which is open...if I just grab the positive wire I wont get shocked, but if I touch a metal piece, will that shock me? Or does the circuit need to close with the same negative terminal?

6. What if Im on an AC circuit and I grab the neutral, then Im pretty sure ill get shocked if I touch a metal part because that completes the circuit with ground, right?

Comments

  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: Basic Electricity Questions
    quique wrote: »
    1. This would mean the phrase "you've got 120V coming out of that wall socket" is conceptually incorrect. Voltage would be the potential energy between two points, such as a waterfall having potential energy from its top (at say, 100m) and its bottom (at 0m). And thus voltage is a differential of electric energy basically between a non live conductor or earth and a live conductor carrying a potential of 120V?

    You definitely do not get "120 Volts coming out of that wall socket". What you have is an electrical potential of 120 Volts at the wall socket. No circuit = no electron flow. The potential is its ability to overcome resistance in a circuit, and air is almost infinite resistance. Lightning can jump through it because its electrical potential is millions of Volts.
    2. In terms of a solar panel V-I graph such as this:
    as voltage, that differential, gets bigger, current begins to decrease until it gets so big that no current can flow anymore. Until it reaches an open circuit voltage which is analogous to opening up a circuit where no more current flows.

    You've just switched gears from a Voltage-based power source to a current-based power source. Assume full illumination of the panel. With zero resistance on the output the panel will produce full current (Isc) at essentially zero Volts. As the resistance increases the Voltage will come up and the current will remain steady until that maximum power point of Imp * Vmp. As resistance increases further Voltage will continue to climb and current will fall until resistance is "infinite" and Voltage is at Voc at which point current is zero. This is the nature of a PV.

    If it were a battery it would try to maintain V against varying resistance and allow current to change. But since no Voltage source can be truly constant over infinite R range the Voltage will fall as current climbs. Believe it or not, batteries have an I*V curve as well and have a point of maximum beneficial power output. This is why they have different total capacities depending on the rate of current draw.

    5. On a related note, if Im on DC circuit which is open...if I just grab the positive wire I wont get shocked, but if I touch a metal piece, will that shock me? Or does the circuit need to close with the same negative terminal?

    If you do not complete a circuit you do not get shocked, regardless of current type or Voltage potential. The trouble is that the higher the Voltage the more likely you will complete a circuit, and you can't always be sure you aren't a part of the current path.

    You can stick your fingers on a car battery and feel nothing, because the Voltage potential is too low to overcome the resistance in your body. By the same token it has enough current potential to weld steel. On the other side is an electric fence charger which is high Voltage and you will feel it, but it doesn't have enough current potential to harm you.
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