P=V*I , P=I^2*R & P=V^2/R why?

Question

They are according to ohms'law:
I have a kettle watt=2400w , resistance=25ohm:
I boiled it with voltage 240v ,I calculate from the 3 equations above & I get the current=10A.
If I boiled it with voltage =120v, accoding to P=VI ,Voltage is inversely propositional to current I=2400w/120v and I supposed to get 20A. But what I get by measuring the actual current is only about 4.8A.WHY???

Answer 1

You are missing the basic law (to use your nomenclature) V = I * R

Simply substitute this equation into that first one for power and you can see how you get the other 2.

Also remember that power is the dependent variable. So when you calculate with varying conditions you can't hold P constant.

Answer 2

Datas are known:

You have a kettle with these specifications:
P = 2400 W
R = 25 ohm

Do you know what those mean?

P = (V*V)/R

V*V = P*R
= 2400*25
= 60000

V = 244.9489 Volts

So you have to boiled your kettle with 244.9489 Volts (245 Volts), rather than 240 Volts (but since the house's voltage source is not exactly constant, it's acceptable to give your kettle V=240 Volts, normally your electricity swings around 230 Volts up to 250 Volts [based on Country's electricity regulation, it can be different to each country] )

And for this normal process (properly process) you'll find:

I*I = P/R
= 2400/25
= 96
I = 9.797959 Amperes

(Did you found it, I = 10A, right?, it's very close, and your true Voltage source is: Vreal = 10 * 25 = 250 Volts !, remember that your voltage source is not exactly constant, as i mentioned it above)

and

How about if you use V = 120 Volts, remember, your Kettle NEEDS V=240 Volts, and you "FEED" it just a half!

Without a deeper calculation, you will find
I = a half of properly Ampere

or

I = 9.797959/2
= 4.898979 Ampere

Is it Matching with your finding?
(Did you found 4.8 A ?)

I hope this mind setting helps.

Answer 3

The element of the kettle is rated at 2400W, but it doen't mean is powered up to that value, hence a difference in the current values.

The first situation:
P = VI
I = P/V
I = 2400/ 240
I = 10 A.

The second situation:
P = VI
P = V(V/R)
P = 120(120/25)
P = 120*4∙8
P = 576w

P = VI
I = P/V
I = 576 / 120
I = 4∙8 A

Answer 4

Mainly because the basic equations are
P= V * I and I= V/R , but the kettle the SAME resistance with the lower voltage as it does with the higher voltage, so I will be 1/2 at 120 V that it was at 240 amps. The kettle doesn't compensate for the lower voltage by cutting its resistance in half.

Answer 5

Resistance generally increases with temperature, so the working resistance will be higher than what you measure at room temperature. Also, check the actual voltage drop across the kettle; depending on your power circuit you could be losing power elsewhere. Also, the power consumption will vary with voltage. At 120V you should expect I=V/R = 4.8A - which is what you got.

Answer 6

This is not uncommon. With such a small resistance heat will be a big factor, it will start out low and get higher as the kettle watt gets hotter. There is also the question of the accuracy of measurements and equipment. But trust the equations, they are correct.

Answer 7

the assumption that it will have same power is what messed you up...
the real constant is the resistance which is 25 ohms

proabably the power rating was based on 240V
current is approx 10A

with
120V and 25 ohms given
current is approx 5 A

which should be about right

Answer 8

how did you measure the current???