Advice please re 12V 50W water heater and 12V 3A power pack

[flockandroll]

I need to do a science experiment like this:

Measuring the temperature of melting ice experiment - Temperature changes and energy - Edexcel - GCSE Physics (Single Science) Revision - Edexcel - BBC Bitesize

https://www.bbc.co.uk/bitesize/guides/zpjpgdm/revision/8

 

I have a 12V power Supply that is rated at 3 amps.

I have a miniature water immersion heater that is “12V, 50W”.

In terms of current, voltage and power, can I safely use the power supply to power the heater? (In terms of wires and connections it is fine.)

 

When I was much younger, I used to think that somehow, “Oh no, 50W at 12V, it will draw over 4 amps to get that 50W at 12 volts, so something will overheat and go pop or worse”.

But now I also think, “There’s only a maximum of 3 amps available, and the heater won’t be able to magically get more than 3 amps, so all that will happen is the heater will run with only 36W, so it will not get as hot, but everything will be safe, and nothing will break”

 

Which view is correct?

The circuit will look something like this:

 

Thanks!

[Edwin_m]

If the power supply is properly designed then it will cut out rather than suffering any damage,, but you won't be able to run the heater.  

[rab]

Your statement about requiring 4 amps is correct.

The power supply is not sufficient, and either

as  @Edwin_m has said  the supply will cut out,

or it will "go pop"

 

Edited April 8 by rab

[Michael Hodgson]

Personally I'd prefer to put the ice into a G&T, relax and drink it slowly while allowing the ice to melt at room temperature.😋

[flockandroll]

Thanks.

So I guess I need to upgrade my power supply.

Mathematically 12A 5A 60W should be sufficient. Will it be sufficient in practice?

https://www.amazon.co.uk/COLM-Adapter-100-240v-Switching-Adaptor-12V-5A/dp/B07HRLGV3S/ref=pd_lpo_sccl_3/261-2754844-5307136?pd_rd_w=TiIAQ&content-id=amzn1.sym.448aab69-6fff-42b9-9d7a-f6325052a6f6&pf_rd_p=448aab69-6fff-42b9-9d7a-f6325052a6f6&pf_rd_r=2KTQJSYVF2SR9F4FEBYX&pd_rd_wg=p3WPg&pd_rd_r=14150e46-7d8e-48c1-8fb9-7a28f44b9e0a&pd_rd_i=B07HRLGV3S&th=1

 

The next one up is 12V 10A 120W, which seems rather a lot of power!

https://www.amazon.co.uk/COLM-Switching-100-240v-Transformers-Accessories-12V-10A/dp/B07J43TMYW/ref=pd_ci_mcx_pspc_dp_d_2_t_1?pd_rd_w=8VGBm&content-id=amzn1.sym.8db49f41-2d78-4c1c-bc3b-fb9e44b9efb4&pf_rd_p=8db49f41-2d78-4c1c-bc3b-fb9e44b9efb4&pf_rd_r=2W566NZ5ERGF892C0ARD&pd_rd_wg=H7Ki9&pd_rd_r=c2f02bc8-f47f-4ec9-af30-29f421ae9698&pd_rd_i=B07J43TMYW

 

Both come with a nifty adapter at the DC end so you can easily connect + and - wires, and I would hope to get some modelling use out of it at some time.

 

So really, I just want to make certain in the real world 12V 5A 60W DC supply will be ok for the 12V 50W heater (it's 20% extra so it must be?). Thanks

[kevinlms]

5 minutes ago, flockandroll said:

Thanks.

So I guess I need to upgrade my power supply.

 

 

Both come with a nifty adapter at the DC end so you can easily connect + and - wires, and I would hope to get some modelling use out of it at some time.

 

 

Maybe you will, maybe not. Both items are currently unavailable. At least to me in Australia!

[meil]

57 minutes ago, flockandroll said:

Thanks.

So I guess I need to upgrade my power supply.

 

Not necessarily - you could put a resistor in series with the heater to work on your power supply. Of course your heater wouldn't be 50W any more but it might be warm enough.

[flockandroll]

Quote

Not necessarily - you could put a resistor in series with the heater to work on your power supply. Of course your heater wouldn't be 50W any more but it might be warm enough.

 

I did wonder, but is it really that simple? And what would the maths be?

[flockandroll]

50W with 12V would be 4.17A

 

R =V/I

 

so R = 12/4.17 =2.88 ohms

 

But I need to be a little under 3A, so total R = 12/just less than 3 = just over 4 ohms,

 

so I would need to add a resistor of about 1.25 to 1.5 ohms in series with the water heater.

 

Is any of that correct?

I can't remember what resistors I have...

[flockandroll]

I have a 100 ohm variable resistor, could I use that and an ammeter to check the current is under 3A ?

Hopefully the heater will still put out enough heat!

[Pete the Elaner]

8 minutes ago, flockandroll said:

50W with 12V would be 4.17A

 

R =V/I

 

so R = 12/4.17 =2.88 ohms

 

But I need to be a little under 3A, so total R = 12/just less than 3 = just over 4 ohms,

 

so I would need to add a resistor of about 1.25 to 1.5 ohms in series with the water heater.

 

Is any of that correct?

I can't remember what resistors I have...

 

It looks right, but the resistor would need to dissipate a lot of power.

 

V=IR

P = VI = IR*I = I^2 * R

R = 1.5 so R^2 = 2.25

I = 3

2.25 *3 = 6.75W. That is a big resistor, or a matrix of smaller ones. Either way, it is a waste of power.

[phil-b259]

7 hours ago, flockandroll said:

I need to do a science experiment like this:

Measuring the temperature of melting ice experiment - Temperature changes and energy - Edexcel - GCSE Physics (Single Science) Revision - Edexcel - BBC Bitesize

https://www.bbc.co.uk/bitesize/guides/zpjpgdm/revision/8

 

I have a 12V power Supply that is rated at 3 amps.

I have a miniature water immersion heater that is “12V, 50W”.

In terms of current, voltage and power, can I safely use the power supply to power the heater? (In terms of wires and connections it is fine.)

 

When I was much younger, I used to think that somehow, “Oh no, 50W at 12V, it will draw over 4 amps to get that 50W at 12 volts, so something will overheat and go pop or worse”.

But now I also think, “There’s only a maximum of 3 amps available, and the heater won’t be able to magically get more than 3 amps, so all that will happen is the heater will run with only 36W, so it will not get as hot, but everything will be safe, and nothing will break”

 

Which view is correct?

The circuit will look something like this:

 

Thanks!

 

The rating of a device relates to its maximum power output!

 

Just because a electric heater may kick out 4W doesn’t mean it has to do that!

 

Given power is a function of Volts multiplied by current the simplest way is to add resistance into the circuit - this will (depending on the configuration used) reduce the voltage or the current being supplied to the heater and consequently reduce the power consumed to less than 4W.

 

Similarly although a power supply may be rated at 3W - that relates to the maximum power it can provide.

 

If the load connected is less than 3A then it will not have a problem supplying power.

 

So if you add sufficient resistance in the circuit and your heater now only draws 2W then both your 3A power supply and 4W heater will function perfectly happily together.

[Edwin_m]

You could just use a car battery, as these can supply huge currents.  According to Wikipedia the cigarette lighter socket is likely to be rated at 10 amps.  

 

 

[flockandroll]

Quote

2.25 *3 = 6.75W. That is a big resistor, or a matrix of smaller ones. 

 

Thanks.

Thanks the 100 ohm variable resistor I have is 5W so not quite up to the job.

 

I guess the bigger power supply will be easier.

[flockandroll]

6 minutes ago, Edwin_m said:

You could just use a car battery, as these can supply huge currents.  According to Wikipedia the cigarette lighter socket is likely to be rated at 10 amps.  

 

 

 

Thanks, but that is not the solution for me...

[30801]

9 hours ago, Edwin_m said:

If the power supply is properly designed then it will cut out rather than suffering any damage,, but you won't be able to run the heater.  

 

It has a 'circuit safe running safely tube' so you're all good.

 

[kevinlms]

17 hours ago, phil-b259 said:

 

So if you add sufficient resistance in the circuit and your heater now only draws 2W then both your 3A power supply and 4W heater will function perfectly happily together.

Except the water will never get hot! Maybe it will be enough to melt ice, which IIRC, was the point of the experiment.

[Edwin_m]

17 hours ago, 30801 said:

 

It has a 'circuit safe running safely tube' so you're all good.

 

I think it's actually a "Circuit not working stoply fuse, when blow no can replace"

[kevinlms]

1 hour ago, Edwin_m said:

I think it's actually a "Circuit not working stoply fuse, when blow no can replace"

Chinese English, for when the smoke escapes, it's f****d!