I will try to ask as intelligibly as possible, please bear with me.

I understood that P = VI which is the power to be transmitted, and V = IR is a voltage loss which must be overcome by transmitting at high voltage and low current.
Now, a transformer works on the principle that it transforms power (ideally with no loss) from one circuit to another. So, a circuit with 10000V and xA will convert to 230V and yA, to retain the same amount of power transmitted.

The real questions:
1] How do two different rated components work from the same line? Eg a 100W bulb and a 40W fluorescent tube?
2] A mobile phone charger uses only 5V and provides a low current to charge the phone. But it has a transformer to step down the voltage from 230V to 5V. Does that not increase the current as per P = VI and transformer principle V1I1 = V2I2?
3] With the understanding that somehow the current passing is decreased as well as the voltage in a mobile charger, where does the extra power go if it is not transmitted to the phone?

  • 3
    \$\begingroup\$ Devices draw power, rather than it being "pushed". A circuit with xA rating is a maximum. \$\endgroup\$ – pjc50 Aug 10 '16 at 13:36
  • \$\begingroup\$ for your first question, P = V*I so what does change for a 100W bulb and a 40W tube when they have the same V? So how can the circuit look like? \$\endgroup\$ – Eggi Aug 10 '16 at 13:37
  • \$\begingroup\$ Your mobile phone may take 1A at 5V from the charger. Using the transformer principle, (which works in both directions) you can see that it takes a tiny fraction of 1A from the 230V supply. \$\endgroup\$ – Brian Drummond Aug 10 '16 at 13:41
  • \$\begingroup\$ @BrianDrummond: My question, how does a ...component "choose" to take a tiny fraction of y Amps at 230V. What controls the current flowing into different components? \$\endgroup\$ – Spandan Aug 11 '16 at 5:55
  • \$\begingroup\$ @pjc50: So, if devices draw power rather than it being pushed, then what happens to the "energy" on the main line that is being continuously transmitted from the powerplant? Because a powerplant is ideally always generating electricity, and putting it out on the line, so if there happens an excess of supply for a city, what happens to the excess energy? \$\endgroup\$ – Spandan Aug 11 '16 at 6:01

I'm going to offer a slightly different tack than the other responses, as you already have enough there to work from. And perhaps a different approach might also help a little. (Or make things worse.)

You may not understand the difference between having the capacity to do something doesn't mean you are actually doing that something. You might have the capacity to lift 100lbs, but that doesn't mean you are always in the process of lifting 100lbs. You might be lifting only 1lb, at the moment.

You bring up a transformer almost immediately. So, let's take that. Buying and using a transformer that has the capacity to handle 100W of power, with its primary side designed to accept 117VAC @ 60Hz and its secondary side designed to deliver 12.6VAC @ 60Hz, doesn't mean that it necessary MUST consume 100W of power at the primary and then magically and forcefully deliver 100W of power at the secondary. It merely means that it may be able to deliver up to 100W, if the secondary load requires it. Not that it must do so if the secondary load isn't present or doesn't require it.

Power can be considered, as you noted, the combination of two things: I and V, or amps and volts, multiplied together. This is like the area formed from the two sides of a rectangle, though.

Most practical systems are designed around setting a voltage (one side of that rectangle) and then allowing the other side (current) to be flexible and to adjust itself. The current is compliant to the area requirement, while the voltage is not compliant. So a device is hooked up to a voltage source, which is not compliant, and draws what current it requires, which is only limited by the ability (compliance) of that voltage source. The voltage source will have a current compliance figure, which is a maximum that a device can reasonably require.

A device hooked up to the voltage source may require a certain power. This means a certain area, in effect. The voltage source is fixed and this sets one side of that rectangular area. And to achieve the needed area in the rectangle, as required by the device, the current adjusts itself (assuming that the voltage source is sufficiently compliant to do so) so that the required area (power) is achieved.

Some power supplies may have their power set in stone, too, as well as their voltage. In such cases, they dissipate as heat what isn't used by the device or load, while performing their other job which is to keep the voltage well regulated. Some power supplies adjust their own loading on their own source automatically, so that they don't dissipate so much as they might otherwise have to. (They may also temporarily store some energy inside themselves, holding it for later use a short moment later.)

There are cases where a power supply doesn't regulate the voltage, but instead regulates the current. Certain kinds of welding, such as SMAW, prefer to use constant currents. So the current is regulated (fixed) and the voltage is the compliant value (which must be within the power supply's compliance rating for voltage.) So, while constant current supplies are not as commonly found, they do also exist.

But power is like area in this sense, with voltage as one side and current as the other side of the rectangle.

Reactive components complicate the understanding a little, because they can temporarily store energy in one of two ways: a magnetic field or an electric field. An inductor stores energy in a magnetic field and a capacitor stores it in an electric field. These temporary sources of energy can be added into, or subtracted from, at different times. These can be used to advantage in a circuit. It's even possible to transfer energy between a capacitor (energy stored, showing as a voltage across its terminals) and an inductor (energy stored, showing as a current through its terminals.) And if they were perfectly efficient in transferring energy back and forth, they could keep doing that transfer between them, back and forth, for a very, very long time.

EDIT: I suppose it may help for you to think of voltage as equivalent to the air pressure presented by an air compressor. If it is at 150psi, that's a certain pressure similar to volts. You can now hook up different attachments, in order to use that pressure to do work. Some of these, like an orbital sander, will permit a lot of flow (cubic feet per minute, or CFM) in order to operate the sander correctly. The design of the orbital sander determines the needed CFM compliance of the air compressor system. And in fact, a lot of air compressors won't properly operate an orbital sander continuously because they can't keep up with the CFM required. They don't have the capacity. Think of the CFM as current, here. Then air pressure is like voltage. Different devices will be designed to use different CFM in order to do their work. Ones requiring less CFM will have higher resistance to the pressure they are exposed to. Ones requiring more CFM will have lower resistance to the presented pressure. These resistances are designed into the tool. And they are usually designed to operate with a specific pressure and CFM rating. And you need to purchase an air compressor that can support the CFM rating at the indicated pressure.

I suppose you could imagine also that the operating power, in this scenario, is the air pressure times the CFM being used while operating the tool. The compliance of the air compressor is the average air pressure it can sustain times the sustainable CFM it can simultaneously deliver, if asked. More sustainable CFM (for a given sustained psi, such as 90psi) means a more powerful unit. (And a more expensive one.)

Many air compressor attachments can still work at widely varying pressures, if poorly at times. Electronic circuits usually depend a little more on a specific voltage level, by comparison. But the basic idea generally holds and some electronic devices (such as the charger for my Nikon Nivo 5C Pro, which accepts any AC voltage from 100VAC to 240VAC) can indeed still operate well over a wide power supply voltage range. So perhaps the analogy still holds, even here.

Finally, consider using just a small nozzle to blow sawdust off of your table saw. That nozzle works great at 90psi, just as the orbital sander does. But it uses a lot less CFM. It has more resistance to the pressure and let's far less air escape per unit time. The air compressor has nothing to say about that, too. It is up to the tool to decide.

  • \$\begingroup\$ ok this really is helpful, especially the difference of capacity vs actual work. However, I have a secondary question, you mention that devices draw according to their needs, so what determines a need? Why doesn't a 230V line push 230V*yA (y is equivalent of what is transmitted from power plant) of power into a component and blow it up? \$\endgroup\$ – Spandan Aug 11 '16 at 6:15
  • \$\begingroup\$ Because voltage is the push, not power. Remember, power is area. Voltage is just the measure of one side. Just because the source supply has the capacity for a certain area, doesn't mean it has any more push (voltage) to achieve it (a longer "voltage" side, let's say.) The loading device can resist that voltage and thereby limit the area being supplied by it. \$\endgroup\$ – jonk Aug 11 '16 at 6:26
  • \$\begingroup\$ so every device has resistance, ok makes sense. So does a CFL (23W) have more resistance than a bulb (100W) to draw less power? Does a 5W mobile charger have even more resistance than a CFL? But a bulb is hotter than all of these... and heat is a measure of resistance. What am I missing? How does a component choose to draw limited power from a line which wants to push more power? \$\endgroup\$ – Spandan Aug 11 '16 at 7:02
  • 1
    \$\begingroup\$ In effect, yes. An important way to reduce the average power consumption of a device is to increase its resistance against the voltage pressure applied to it, while at the same time still doing the same designed tasks required of the device when provided access to said voltage. Sometimes, devices just shut themselves off completely and that greatly increases their resistance... for a time. (Like a refrigerator does, by turning on and off, periodically.) In general, and on average, if a device uses fewer watts given access to the same voltage supply, then they have higher average resistance. \$\endgroup\$ – jonk Aug 11 '16 at 7:10
  • \$\begingroup\$ I'll add a note to my answer that may help a little. \$\endgroup\$ – jonk Aug 11 '16 at 7:25

The real questions: "1] How do two different rated components work from the same line? Eg a 100W bulb and a 40W fluorescent tube?" ------ House hold devices are connected in parallel. So, appliances with variable power rating (100W and 40W in your case) working at 230V would draw different currents (i.e. 100W/230V and 40W/230V in your case respectively).

"2] A mobile phone charger uses only 5V and provides a low current to charge the phone. But it has a transformer to step down the voltage from 230V to 5V. Does that not increase the current as per P = VI and transformer principle V1I1 = V2I2?" ------ A mobile phone charger doesnot have a transformer. It has a regulator instead that converts 230V ac to 5V dc. It depends on the charger rating howmuch can the current be and the battery rating as well how fast it can charge.

"3] With the understanding that somehow the current passing is decreased as well as the voltage in a mobile charger, where does the extra power go if it is not transmitted to the phone?" ------ There is always some loss in any kind of conversion even in charger. But the conversion you are considering is incorrect. The loss depends on the rating of the mobile charger. If the charger rating is 1A, then it can support only 1A of current passing through it. An if this changer is used to charge a 1000mAH battery then it would 1hr to charge it.

Hope it clears your doubt.



How do two different rated components work from the same line? Eg a 100W bulb and a 40W fluorescent tube?

P = V I therefore 100 watts = 230 V x I OR I = 0.435 amps

P = V I therefore 40 watts = 230 V x I OR I = 0.174 amps

A mobile phone charger uses only 5 V and provides a low current to charge the phone. But it has a transformer to step down the voltage from 230 V to 5 V.

A mobile phone charger produces 5 V and let's say the phone takes 1 amp. That's a power of 5 watts. That power has to come from somewhere i.e. the 230 V AC supply so using the formula used above....

P = V I therefore 5 watts = 230 V x I OR I = 0.022 amps

Of course there will be a little bit more power taken due to imperfect 100% efficiency.

where does the extra power go if it is not transmitted to the phone?

Power taken is what is needed to be delivered (as an output) plus any power needed to make the process work.


1) in your Q the two devices are connected to the same power-source parallel, so they "feel" the same voltage. If they have different (inner)resistance the current passing through each device is different. The higher the resisance the lower the current: current [A] = voltage [V] / resistance [Ohm]

2) yes, the current on the 5V side is higher (as you assumed due to formula - which is correct). It is NOT the same current on both sides, so current on the 230V side is much lower. e.g. If Voltage would be half, then current is dubbled,...by the way a moble phone charger does have a transformer (with high fequency) in it.

3) example (idealized - means no losses): ratio of voltage on secondary side (= 5V) compared to voltage on primary side (= 230V) is 5 / 230 = 1/46. So then current on secondary side is 46 times higher than current on the 230V (primary) side. If seconary current is 1A then on 230 V side ist must be 1/46 A.


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