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Discussion in 'General Electronics Discussion' started by Agriias, Oct 16, 2014.

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  1. Agriias


    Oct 16, 2014
    Hi guys,

    I have some questions that I am struggling to understand. Due to my situation I don't have much chance to experiment much and am stuck doing a lot of book learning. I understand how to solve for basic circuits.

    Here are my questions:

    1.) House circuits have 120v 15A, however if I divided 120v by a load of 2ohm I would get 60AMPS. Obviously this is not the case. Do wall sockets have a natural resistance of 8 ohm thus giving 15A? How is it that you can switch to a 20amp socket? How many amps are actually running through a circuit? I understand most houses have 100amp breakers and sometimes larger, is there a transformer in it to drop the amps?

    2.) what is the voltage and amps of a typical residential powerline? How about the large lines? Is there a resource on the site with this information?

    3. I am sort of stumped by the idea of total circuit current. For instance, if I have a basic circuit with a parallel section of 3, 2ohm resistors connected to 12 V source, they would be equivalent to 2ohms. Each would have its own branch current as well of 2amps adding to total I of 6a. Here is where I am confused, if I was to connected another path after the furthest most resistor , but before it meets at the shared node where all branch currents combine, would any load I hook up to that have only the individual branch current pushing through it?

    I think what I am struggling with in any circuit design is whether to create essentially numerous parallel circuits that split right off at the source and that only recombine after their loads function has been accomplished or whether it is useful to hook different components up in series.. The entire idea of branch current is confusing. do we determine total circuit current just as a mathematical device to check our work or does it help us determine the actual voltage and current present at certain points? can the voltage and current be different in the same circuit?

    Example, if I had a 12v source connected to a 2ohm load there would be 6amps. This would be 72 watts. However, if I added another 2ohm resistor in series I would now have 3amps and only 36 watts. However the voltage drop at each would be equally 6v. in the first example all 12v were applied to the first resistor, but how many volts and amps would read if you measure after that resistor? How many after the second resistor in example 2?

    4. Wall worts confuse me.. How is it that they are dropping the voltage and current? A giant resistor? It cant be an inductor in there because where did all the extra voltage and current go?

    5. in a transformer I understand that the ratio of primary to secondary winding effects the relationship because voltage and current.. however, what effect does the size of the magnetic wire used in the turnings play? besides that smaller wire allows for more turns? if I had two transformers with the same ratio but with different gauges what would be the practical differences?
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Hi Agriias and welcome to Electronics Point :)

    Amps (current) is a FLOW. When nothing is plugged into a wall socket, no current flows at all. The 15A specification is just a safety limit. It's telling you that you can draw UP TO 15A of current from that socket.

    If you draw more than that amount of current, the socket and/or the wiring may overheat, and a fuse should blow.

    Current is a FLOW, but voltage is a DIFFERENCE, like a distance. It is always measured BETWEEN two points. With a wall socket, and any power source, the voltage is (in principle) constant.

    You refer to resistance, so I guess you know Ohm's Law: I = V / R
    I is current, in amps (A);
    V is voltage, in volts (V);
    R is resistance, in ohms (Ω).

    A typical 1 kW bar heater is made from resistive wire, e.g. nichrome wire. For use at 115 VAC, the element will have a resistance of roughly 13Ω. When you plug the heater into the wall, current flows through the heater's power cable, through the switch, and through the element. The amount of current will be:

    I = V / R
    = 115 / 13
    = 8.8A approximately.

    This current is the same in all parts of the series circuit - that is, the wall wiring, the socket contacts, the plug pins, the power cable, the switch, and the heating element. The same current flows through all of those points, in a big loop.

    The reason that the element gets hot, but the other parts don't (though they may get warm) is that heat is produced when POWER is dissipated, not just when current flows. Power is dissipated when there is a current flow AND a voltage drop. In fact the formula for power is just those two quantities multiplied together. This is called the Power Law:

    P = V × I
    P is power dissipated in a part of a circuit, in watts (W);
    V is the voltage ACROSS that part of the circuit, in volts (V);
    I is the current flowing THROUGH that part of the circuit, in amps (A).

    These two formulas can be combined in various ways. We can convert Ohm's Law from I = V / R to V = I × R, then substitute it for V in the Power Law. That gives us:
    P = V × I
    P = (I × R) × I
    P = I × R × I
    P = I2 × R.

    In the heater example, I (current) is about 8.8A, so I2 is about 77. We can then say that each part of the circuit will dissipate an amount of power equal to 77 multiplied by its resistance in ohms.

    Most of the parts of the heater circuit are carefully designed to have low resistance. The wire is made of lots of strands of good-quality copper (unless it's a Chinese "firestarter", but that's a different subject), the switch uses copper and possibly gold plating, and so on. If the power cable's resistance is, say, 0.03Ω then it will dissipate (77 × 0.03) = 2.3 watts over the whole length of the cable. Not enough to make it more than a few degrees warmer than the ambient temperature.

    However, the element has a resistance of about 13Ω and will dissipate (77 × 13) = 1001 watts, i.e. 1 kW. This energy will be converted to heat (and a bit of red light).

    Please go over that description carefully until you understand the characteristics of voltage, current, resistance, and power.

    You will also need to learn about RMS and the differences between DC and AC but that can wait for another time :)
    Probably varies from country to country. Here in New Zealand, neighbourhood wiring is normally 11 kV. At other parts of the electric power distribution network it is 33 kV and 110 kV. The HVDC (high voltage DC) cable between the North and South islands operates at 500 kV!

    Try Googling keywords like electricity distribution network urban feeder voltage.
    Remember that current is a FLOW THROUGH a point in a circuit. Voltage is a DIFFERENCE BETWEEN two points in a circuit. If you connect two resistors in parallel, they will have the same voltage across them. If you connect two resistors in series, they will have the same current flowing through them. In a series circuit, current at all points is the same.

    If you connect, say, one 2Ω resistor across a 12V DC source, the current flowing through that resistor will be, from Ohm's Law, 6A. Now if you connect another 2Ω resistor across that power supply, 6A will flow through it as well. Now if you connect a third 2Ω resistor, another 6A will flow through that one.

    Look at the circuit you have and you'll see that the three resistors are in parallel. Each resistor is connected across the power supply, so each one has 12V across it, and because each resistor is 2Ω, each resistor has 6A flowing through it.

    The power supply is providing three lots of 6A, one for each resistor, so the power supply is actually providing 18A in total to the total combination of paralleled resistors.

    If you work backwards, rearranging Ohm's Law to R = V / I, the calculated resistance is 12 / 18 or 0.667Ω. THIS IS, in fact, the combined resistance of three 2Ω resistors connected in parallel.
    I hope I've explained that well enough above. Always remember that voltage is a DIFFERENCE BETWEEN two points in a circuit, and current is a FLOW THROUGH a specific point in a circuit.
    Yes, you're right so far. If you connect another 2Ω resistor in series with the first 2Ω resistor, the total resistance is now 4Ω so the current will drop from 6A to 3A and the voltage will be shared by the two resistors.
    "measure after that resistor" doesn't mean anything.
    With two 2Ω resistors in series across a 12V supply, 3A will flow through both resistors and each resistor will have 6V across it, so each resistor will dissipate 18W.
    They have a transformer. Either a heavy iron-cored one that operates at line frequency, or a small ferrite-cored one that operates at a higher frequency that is generated by some circuitry inside the wall wart. The latter type are called switching supplies.

    Both types step the voltage down, and in the process, they step the current up. For example, a 12V wall wart powering a lamp that draws 1A will draw a lot less than 1A from the wall socket.
    Yes, thinner wire allows for more turns. It also has a higher DC resistance and a lower current handling capability. So a step-down transformer has a primary made of thin wire, where the current is relatively low, the voltage is high, and many turns are needed, and a secondary made of thicker wire, because the transformer's output current is relatively high, and not so many turns are needed.
    Last edited: Oct 16, 2014
    Supercap2F, hevans1944 and davenn like this.
  3. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    I think @KrisBlueNZ answered your questions very well. I have some additional comments and the following observation: it appears as if you are trying to learn electrical circuit theory starting with a basis of house wiring and low-voltage battery operated equipment. That is an excellent approach, but be careful around things powered by house wiring. Remember it has the full faith and credit of your electrical utility backing it up!

    Yes it is! Kris explains why.

    A typical residential power line in the United States is three heavy-gauge wires connected from a transformer to the house. The wires will be sized large enough to carry the rated maximum current of the circuit breaker panel, to which they are connected. The purpose of the circuit breaker panel is to distribute the three wires to loads throughout the house via circuit breakers in series with the loads. The circuit breakers protect the house wiring if a fault (short circuit or other overload) should occur.

    Other houses will probably share the transformer and have their own three wires connected in parallel with yours at the transformer. In my neighborhood the primary of the transformer (called a "pole pig" if it is mounted on a pole near the house) has 7,000 V AC exciting it. The secondary winding is 240 V AC and is center tapped to create a so-called grounded neutral. The neutral in my house is bonded to a cold-water pipe coming out of the concrete floor of my basement.

    There is 120 V AC between each end of the transformer secondary winding and the neutral. Most household loads, such as lamps and small appliances, are connected between the neutral and, through a circuit breaker, to one of the other two transformer secondary wires to provide 120 V AC. Heavy loads, like an oven or a clothes drier connect between both of those two secondary wires, through 2-pole circuit breakers, and will have 240 V AC supplied to them.

    Again, @KrisBlueNZ explained this. It is probably the most difficult concept for a beginner to understand without using Ohm's Law and Kirchoff's Law to actually analyze a circuit, plugging in numbers for voltages and resistances and solving for the resulting currents. The math is essential to understanding, not just a "device" to check your work.

    Actually there IS an inductor in there, either a small transformer as Kris explained, or much more complicated circuitry involving inductors, capacitors, resistors and fast-switching semiconductor devices. The important thing to realize is this: if a wall wart were 100% efficient, the product of the voltage in times the current in (=power in) would equal the product of the voltage out times the current out (=power out). Real wall warts are NOT 100% efficient: the power in will always be slightly greater than the power out. That is why wall warts get warm after operating awhile. The heat comes from the extra power consumed because they are not 100% efficient.

    Transformers appear simple, but they are not. An ideal transformer behaves like an ideal wall wart (except transformer outputs are always AC while a wall wart can provide either low-voltage AC or DC) so power in equals power out. If the transformer secondary is unloaded (open circuit on the secondary) very little power is drawn by the primary, just the so-called magnetizing current which causes some heating because it flows through the resistance of the wire used to wind the primary winding. This current increases as the load on the secondary increases and the transformer delivers power to the load. For a step-down transformer, the secondary wire is sized to accommodate the maximum current for which the transformer is rated. The practical difference between "two transformers with the same ratio but with different gauges" is the power handling capacity. The power that a transformer can transfer from primary to secondary is mainly determined by the area of the core on which the windings are placed. More power means a larger core is necessary. It is a lot more complicated then that, but all you need to know is this: if the turns ratio is the same, the larger transformer will be able to provide more power to the load and the gauge of the primary and secondary windings will be sized accordingly.
    Supercap2F and KrisBlueNZ like this.
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Great explanation Hop!
  5. Agriias


    Oct 16, 2014
    I'd like to thank you both for your help! I'm sorry to be like the kid in home alone with a thousand questions.. "do these trucks get good gas mileage?"

    So for the "pole pigs" that are 7kV that is 58 times 120volts. Does that mean that the current is 58 times less than whatever your breakers maximum current is? so for 100A breaker 1.72Amps? I know that Voltage and Current are inversely related when transformed, but when the energy is generated say back at the power plant what is the initial voltage/amps? is it 120v/120amps and then stepped up?

    Ok so in series circuits the amps are the same but voltage varies based upon the drops. If you had say two light bulbs in series, but after the first bulb the voltage drop was significant and the second bulb couldn't be powered. what would be the result? would the first bulb light up? Would both not light up since the second load doesn't have enough voltage and its resistance would be high enough to virtually stop the flow of current?

    Ok I got it on the FLOW with current. I get it now with voltage.. I had to kind of think about it in terms of charge and static electricity and then how a magnet can polarize a piece of metal.

    Now I got to figure out series-parallel circuits. I know you essentially solve for them by figuring out the reciprocal values and turning them into a big series circuit to solve for total current. It may be a big confusing now to figure out how a series parallel circuit behaves. I would assume that if you had 3 in series resistors follow by 2 resistors in parallel of different resistances that the branch current would be different, but the voltage going into their shared node would be the same... to find out the voltage drop across the parallel portion of the circuit how would you figure it?

    Thanks again guys I appreciate you taking the time to answer my questions.
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Well, each pole pig will be supplying hundreds of houses, which are connected in parallel so their currents add together. But yes, if your house is drawing 100A at 120V, that will contribute about 2A to the current flowing in the 7 kV feeder.
    No! Think about the big picture here. Say your house is drawing 3 kW. Your suburb might have 1000 homes; that's 3 MW (megawatts). Your town might have 20 suburbs; that's 60 MW. The power distribution system that takes power from where it's generated to where it's used operates with numbers in the gigawatt range!

    To transfer even 60 MW of power through copper wires, you need to use a very high voltage. The reason is the resistance of the wire and the current flowing through it. Remember that P = V × I. Power is voltage multiplied by current. For a given amount of power, lower voltage means higher current, and vice versa.

    A cable carrying 60 MW at 120V would be carrying a current of (60,000,000 / 120) = 500,000 amps! That's a HUUUGE amount of current. With that much current flowing, ANY resistance in that cable is going to dissipate a huge amount of power. Remember, P = I2 × R, and 500,000A squared is 250,000,000,000! So even a piece of huge fat copper cable with a resistance of, say, 0.0001Ω, which is almost a perfect conductor, will dissipate 25 megawatts! It would vaporise in a microsecond!

    To transfer large amounts of power while keeping the current manageable, you have to use higher voltages. A cable carrying 60 MW at 110 kV will be carrying a current of (60,000,000 / 110,000) = 545A. That's still a hell of a lot of current, but nothing like the 500,000A that would be needed if the distribution voltage was only 120V. More importantly, I2 is only about 300,000, not 250,000,000,000, so a small amount of resistance won't cause such a dramatic amount of power to be dissipated in the cable itself.

    So for practical reasons, because even copper is not a perfect conductor, higher power levels normally correspond to higher voltages.
    That doesn't make sense. Here's a better analogy.

    Think of a lamp as a tension spring. One of these:

    DTS illustrative springs.png
    If you hook two identical tension springs together, end-to-end, and stretch them apart to a certain distance, each spring will stretch to half of that total distance. That's how voltage is split when you connect two identical lamps in series and apply a voltage across them.

    This analogy, or model, can be extended to explain current and resistance as well:

    Distance === Voltage (always measured between two points);
    Tension === Current (always measured at a certain point in a circuit);
    Stretchiness === Resistance (a characteristic of a component or part of a circuit).

    I've shown some resistance values next to the springs in the picture above, because resistance sometimes seems like a backwards concept. A thick, strong spring is equivalent to a low resistance - the 10Ω spring at the right. A thin, weak spring is equivalent to a high resistance - the 1000Ω spring at the left. That's why I said that resistance is equivalent to stretchiness - the more stretchy (easily stretched) the spring is, the higher its resistance.

    You can visualise Ohm's Law using this model as well. I = V / R means that you will have more tension (current) if you (a) stretch the spring over a greater distance (higher voltage), or (b) if you use a thicker spring (lower resistance), or (c) both. Ohm's Law also tells you exactly how those quantities are related, arithmetically.

    This simple analogy, comparing a resistor to a spring and equating voltage to distance, and current to tension, allows you to visualise many simple circuit that use resistors (or lamps) in series and parallel. For example, a voltage divider, which consists of two resistos in series, and is widely used in electronics.

    Your two lamp circuit is actually a voltage divider. The total applied voltage is divided in half. Half appears across one lamp, and half appears across the other lamp. But if the lamps weren't identical, you could imagine the lower-resistance one as a thick spring, and the higher-resistance one as a stretchy spring, and connect them in series. Then you would see how the total voltage would be distributed between them. Which one would have more voltage (distance) across it?

    This analogy, which I call the "DTS model" (DTS for distance, tension, stretchiness), also allows you to easily understand the behaviour of capacitors - one of the great mysteries for electronics beginners - and to some extent, LEDs and other semiconductors. I've covered various aspects in other posts here on Electronics Point; do a search for "DTS model" and you'll find them if you're interested.

    The DTS analogy is not well-suited to some aspects of electronics. For example, in the real world, current is the flow of electrons, but nothing "flows" in the DTS model - current is represented by tension, which is static. Also the concept of power dissipation is not modelled. It has its uses, but like all analogies, it is not perfect.

    Another obvious problem with using tension springs to model resistors is that tension springs have a fixed initial length, so if you don't stretch the spring, it still has distance across it, whereas a resistor with no current flowing has no voltage across it. In the picture in your mind, you have to ignore, or subtract, the initial length of the spring!

    Also for completeness I should mention that lamps are not exactly like resistors. But they are close enough for these principles to apply.
    Oh dear! I think you'll find the DTS model is a better way to think about voltage.
    I would like to know whether you can work out the answers to some of your questions using the DTS model. I think if you're mechanically minded and you understand the three DTS quantities clearly, you should be able to imagine resistors hooked in series and parallel, and how voltages and currents will be distributed.
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