# Two thermophores boiling water

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1. Oct 6, 2016

### ChessEnthusiast

1. The problem statement, all variables and given/known data
We are given two identical thermophores. It is known that one of them can boil a liter of water in t = 600 s.
How much time would it take to boil one liter of water if we used two of these thermophores, connected
a) in a series circuit
b) in a parallel circuit
Voltage: 230 V

t = 600 s
U = 230 V
I1 = I2

2. Relevant equations

W - energy required to boil water

W = UIt

3. The attempt at a solution
a) series circuit

Since the thermophores are in a series, each of them will share the same current, thus:
W = (UI+UI)t = 2UIt => t = 300 s

b) parallel circuit

The thermophores are in a parallel connection, which means that they will share the current, and so each of them will be empowered with the current 0.5I

W = 0.5I U t + 0.5 I u T = UIt => t = 600 s

However, I doubt that this solution is valid.
Could you, please, give me some tips or reference how to tackle this problem?
Thank you very much in advance.

2. Oct 6, 2016

3. Oct 6, 2016

### TSny

Yes, each thermophore in the series circuit will have the same current. But will that current be the same amount of current as when there was only one thermophore in the circuit?

Likewise in the parallel circuit you will need to consider whether or not each thermophore in the parallel circuit has the same current as in the original circuit with just one thermophore.

4. Oct 7, 2016

### ChessEnthusiast

These thermophores are identical, and so they have identical resistance, and so if voltage is the same, then the current will also be the same,
right?

5. Oct 7, 2016

### TSny

When the two devices are in series, does each device have a potential difference of 230 V?

6. Oct 7, 2016

### ChessEnthusiast

Hmm..
U1 + U2 = 230 V

U1 = U2

Because the two thermophores are identical, and the Voltage needs to add up
Is that correct?
If so, that would mean that in the series circuit, the Voltage will be
U = 115 V

7. Oct 7, 2016

### TSny

Yes, good. In the series circuit the potential difference across each thermophore will be 115 V.

(Another approach to the problem, which you can use to check your result for the total power, is to replace the two resistances in series by one equivalent resistance and deduce the power for the equivalent resistance.)

8. Oct 7, 2016

### ChessEnthusiast

But now, I am getting a very unintuitive answer.

Since Useries = 0.5U
and
W = (U^2 / R) t

Substitute:

W = Useries2 tseries / R

W = (U^2 / 4R) t

t series= 4WR / U^2 = 4 t

Is it possible that time required by these two thermophores is FOUR times bigger than that required by only one?

9. Oct 7, 2016

### TSny

Is this the heat produced by just one of the thermophores in series or is it the total heat produced by both thermophores together?

10. Oct 7, 2016

### ChessEnthusiast

Ohh, right. It is the heat produced by only one thermophore.
The heat produced by both of them will be:
$$W = \frac{U^2}{2R} t$$
And so
$$t_{series} = 2t$$

Still, am I missing something?

11. Oct 7, 2016

### TSny

That's the right answer for the series case. I don't think you're missing anything.

12. Oct 7, 2016

### ChessEnthusiast

Well, I need to say that this answer is pretty unintuitive.
I would never say that two thermophores will require more time to boil water than only one.

That's the beauty of Physics:)

13. Oct 7, 2016

### TSny

Yes.

Now you can move to the parallel case. The answer you gave in the first post is incorrect.

14. Oct 7, 2016

### ChessEnthusiast

In the parralel case, the total current:
$$I = U(\frac{1}{R} + \frac{1}{R}) = \frac{2U}{R}$$
Hence,
$$W = U I t = U^2 \frac{2}{R} t$$

And finally
$$t_{parallel} = \frac{t}{2}$$

I feel much better with this answer.

15. Oct 7, 2016

Good work!