-1
$\begingroup$

I'm trying to understand a bit about electrical circuits. I've understood the basic equations like V = IR. They are a few things I'm confused about. My understanding is that when you have a circuit that is connected to a battery there has to be some resistor in the circuit of some kind to ensure the current on the wire is small enough to power the device (e.g a bulb).

I'm looking at a power cord (British style one); just the part with three prongs the goes to live, neutral and ground(earth) it isn't connect to any charge or device its just the part of the wire that comes out of the cord. It has some readings on it, namely it states 3A/250V~ which I understand as 3 Amps, 250 Volts, Alternating Current. Is the writing describing that it will allow 3 Amperes to flow through it; if so how is it doing that? Is it using a resistor inside it or what? It also has a clearly visible fuse labeled with the same 3A; but surely if there is no resistor inside the power cord and if I connect the power cord directly to the 240V supply from the mains sockets to a naked bulb for example. Then wouldn't the 3 Amp fuse immediately blow because the bulb doesn't have any resistor in it so it would just try to draw as much current as possible and the fuse on the power cord only allows the 3 Amp current due to the fuse. Is this the right thought process?

My Final query is in regards to why there is writing directly on the power cord saying 3A/250V~ and then the fuse itself also says 3A. I understand the fuse but what does the 3A on the power cord mean; if it's referring to the fuse you could just replace the fuse with a different Amp value one. What am I not understanding?

Edit:

Wikipedia shows that polarity symbols on a charger are supposed to be either negative to positive or vice versa on the symbol. A charger I have shows negative on both sides of the drawing what does this mean; Is that valid? Polarity Symbol

$\endgroup$
  • 1
    $\begingroup$ The bulb is a resistor. $\endgroup$ – Jonathan R Swift Sep 5 at 7:28
  • $\begingroup$ The bulb is a resistor that glows when it gets hot! $\endgroup$ – StainlessSteelRat Sep 5 at 19:43
4
$\begingroup$

My understanding is that when you have a circuit with that's connect to a battery they has to be some resistor in the circuit of some kind to ensure the current on the wire is small enough to power the device (e.g a bulb).

The bulb is the resistor! The power supply (battery) will provide the power the load requires subject to the battery's capability limits.

I'm looking at a power cord ... It has some readings on it, namely it states 3A/250V~ which I understand as 3 Amps, 250 Volts, Alternating Current.

Correct.

Is the writing describing that it will allow 3 amperes to flow through it; if so how is it doing that?

The wire itself has some resistance. When current flows through the cable there will be some power dissipated in it and this can be calculated from $ P = I^2R $. Cables should be sized for the designed load. Note that the heat goes up with the square of the current so doubling the current gives four times the cable heating.

Is it using a resistor inside it or what?

Definitely not.

It also has a clearly visible fuse labeled with the same 3A; but surely if there is no resistor inside the power cord and if I connect the power cord directly to the 240V supply from the mains sockets to a naked bulb for example. Then wouldn't the 3 amp fuse immediately blow because the bulb doesn't have any resistor in it so it would just try to draw as much current as possible and the fuse on the power cord only allows the 3 amp current due to the fuse. Is this the right thought process?

No. The bulb is the resistor! It limits the current.

Why there is writing directly on the power cord saying 3A/250V~ and then the fuse itself also says 3A.

It's telling you that the cable is rated at 3 A so you should use a 3 A fuse to protect it in the event that it is used to power a larger load (lower resistance). The fuse will then blow to protect the cable.

A charger I have shows negative on both sides of the drawing what does this mean; Is that valid?

No. A plug with negative on both pins will supply 0 V, 0 A and 0 W. (At least the maths would be easy.)


Am I correct in thinking that to change the current you have to change the resistance or voltage?

The relationship is determined by Ohm's Law. $ I = \frac V R $ so changing either V or R will change the current.

The bulb is a resistor and the wire is also a resistor?

Yes. Omni calculator shows that 2 m of 0.75 mm2 cable has a resistance of 46 mΩ.

Meanwhile a 230 V, 100 W incandescent lamp will have an on-resistance of $ R = \frac {V^2} P = \frac {230^2} {100} = 530\ \Omega $.

As you can see the cable resistance is negligible (1/10000) compared to the lamp. The lamp gets hot. The cable doesn't.

3A fuse is there to protect the wire because it can only handle 3A.

OK.

If the bulb is the resistor and for example we know it's resistance is 2Ω and I supply the 2V voltage the current would be 1A.

Your reasoning is correct. Remember that the resistance will increase by a factor of up to ten as the lamp heats up to working temperature.

If a device/bulb specifies its resistance as 2Ω that does not say anything about the current it requires right?

No. The lamp would require at least two parameters - voltage and resistance in your example but voltage and power (W) in real life.

I can't determine what voltage to supply unless given Amps. right?

No. Most devices have a rated voltage and draw whatever current they require. Your cooker will have a clock, several rings, a grill an oven and fan. You supply the right voltage with a cable capable of carrying the worst case current. The required voltage is stamped on the rating plate on the rear of the cooker.


If I have a bulb rated 10V, 40W for example. That implies its working current as 40W/10V = 4A. But I have a power supply of 20V. could I make that supply work with the bulb? Could I add a resistor to the circuit and effectively half the current to around 2A and the resulting Power would stay the same (2A * 20V) = 40W.

No, you've got mixed up here. You need to keep the current the same (4 A) but reduce the voltage by 10 V. So the resistor has to have the same resistance as the (hot) bulb and will dissipate the same power (40 W). Your efficiency drops by 50% (and incandescent lamps are notorious for giving more heat than light).

Basically I'm asking does the bulb(or any device) care about the current or care about the actually power dissipation.

You can't have one without the other. $P = VI = I^2R $ so they're directly related.


I have to keep the current the same why?

The power in the lamp is given by $ P = I^2R $. If you change the current then the brightness of the lamp will change.

The same power is being dissipated.

No. A 40 W, 10 V lamp will have a thicker filament with lower resistance so it passes more current than a 40 W, 20 V lamp. Have a look at the difference between a 60 W house bulb and a 55 W car headlamp.

I understand that the current and power are related but I can have the same power with any number of different values for the current and voltage.

But each bulb is designed for one specific voltage. The thickness and length of the filament determine that. Thin and long for high voltages. Short and fat for low voltages.

4 A * 10 V and 2 A * 20 V give the same power. Then where is the problem?

You didn't calculate the filament resistance required.

I'm assuming power is the work that will be done every second.

Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. It is measured in watts (W).

Energy is power by time and is measured in watt-second (Ws) (which are the same as joules (J)) or for billing purposes in kilowatt-hours (kWh).

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for answering my question. Am I correct in thinking that to change the current you have to change the resistance or voltage? the bulb is a resistor and the wire is also a resistor? 3A fuse is there to protect the wire because it can only handle 3A. If the bulb is the resistor and for example we know it's resistance is 2Ω and I supply the 2V voltage the current would be 1A. Here's another question if a device/bulb specifies it It's resistance as 2Ω that does not say anything about the current it requires right? I can't determine what voltage to supply unless given Amps. right? $\endgroup$ – BrianO Sep 5 at 12:47
  • $\begingroup$ See the update. $\endgroup$ – Transistor Sep 5 at 13:03
  • $\begingroup$ When you said a 230V, 100W incandescent bulb. That is referring to the fact that the bulb wants a 230V supply and will have a 100W/230V (0.43A) current flowing through it. $\endgroup$ – BrianO Sep 5 at 14:52
  • $\begingroup$ You've also said the resistance will increase by a factor of up to 10 as the lamp heats up. Did I miss something or is this just something that happens when you heat up something. It's resistance changes. are you saying that while the lamp heats up the resistance is dynamically changing. Also why up to 10 times. $\endgroup$ – BrianO Sep 5 at 14:54
  • $\begingroup$ All conductors have a temperature coefficient of resistance. Almost always the resistance increases with temperature. See wired.com/story/….. $\endgroup$ – Transistor Sep 5 at 14:59
1
$\begingroup$

A shot of one of my many power "bricks:"

power brick close up

I suspect that you're seeing a typographical error that should have never made it through to production.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for this but I'm more concerned with the initial part of my question. If you could help with that I'd be grateful $\endgroup$ – BrianO Sep 5 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.