# Rated voltages

1. Oct 13, 2015

### crom1

1. The problem statement, all variables and given/known data
Two resistors of equal rated voltages $V_n$, and different rated powers $P_{n1}= 50 W$ and $P_{n2}= 100 W$ are connected in series on source of rated voltage. Find total power of a circuit.

2. Relevant equations $P=\frac{U^2}{R}$, $R=R_1+R_2$

3. The attempt at a solution
I am not quite sure what rated voltage is. I mean if two resistors have same voltage, and are connected in series, shouldn't they have same power? (obviously the word "rated" makes some difference but I don't know how).
My attempt was that using that voltage of a source is sum of voltages of the resistors (which it really is in series ) and using *relevant equations* but instead of getting $P=33.33 W$ I get $P=133.33 W$ . I notice that If $V_{source}= V_{n}$ I would get the right result.
Any help?

2. Oct 13, 2015

### CWatters

If the rated voltage Vn is applied to a resistor it will dissipate the rated power.

Try writing equations for R1 and R2 in terms of Vn and Pn.
Then try drawing the new circuit. The problem says the voltage source is also Vn
How would you normally simplify a circuit with two resistors in series?
Note the problem statement only asks for the total power.

3. Oct 13, 2015

### Hesch

That's generally not correct.
I've had a look at some SMD-resistors ( 0.125W ), and for all resistors Vn = 100Vdc though the resistor values are within a range of 1Ω to 1MΩ.
The Vn value is determined from the max. field strength (V/m) and thus from the mechanical size of the resistor, not from the power.

4. Oct 13, 2015

### CWatters

Agreed. But I think it's the only way to solve this problem.

5. Oct 13, 2015

### Staff: Mentor

Agreed -- it's a strangely worded problem.

@crom1 -- can you upload a scan of the problem?

6. Oct 14, 2015

### crom1

The original problem is not in english. (unless someone here doesn't know croatian I don't see a point in posting it). I don't see a mistake in my translation but it is possible there is a mistake, I will try to ask maybe on some croatian forum to see if the problem makes sense.

7. Oct 14, 2015

### CWatters

Don't go because I think you have the right answer.

This is how I approached it...

P = V2/R
so
R1 = Vn2/50
R2 = Vn2/100

In the new circuit you have R1 and R2 in series with Vn so the new power PT is

PT = Vn2/ (Vn2/50 + Vn2/100)

Vn2 cancels

PT = 1 / (1/50 + 1/100)
= 33.33 W

8. Oct 14, 2015

### crom1

Yes, I noticed that solving it like that, I would get the right answer. But I have question, if the voltage of a source is $V_n$, then how come the voltages on the resistors are also $V_n$, since in series we have that $V_{source}=V_1+V_2$ and in this case it would be$V_n+V_n= V_n$?

9. Oct 14, 2015

### CWatters

That's not correct. In the final circuit the voltage source is Vn (from the problem statement) but the voltage on each resistor is not Vn. I'm not sure it's even possible to work out the voltage on each resistor based on the info given. See next post.

They carefully choose the voltage source to be Vn just so that it would cancel later. It's one of those problems that tries to fool you into thinking you don't have enough information but in reality it's been carefully crafted so that the info you think is missing isn't actually needed.

Last edited: Oct 14, 2015
10. Oct 14, 2015

### CWatters

Actually I think you can write an equation for the voltage on each resistor in the final circuit..

The potential divider rule can be used to write...

V1 = Vn * R1/(R1+R2)...............(1)
V2 = Vn * R2/(R1+R2)...............(2)

We know from the problem statement..
R1 = Vn2/50
R2 = Vn2/100

Substitute into 1...

V1 = Vn * Vn2/50 / (Vn2/50 + Vn2/100)
Vn2 cancel leaving
V1 = Vn * 1/50 / (1/50 + 1/100)
which further simplifies to
V1 = 2/3 * Vn
and so
V2 = 1/3 * Vn

11. Oct 14, 2015

### crom1

One thing is unclear to me.
Why is
$R_1 = \frac{V_n^2}{50}$
$R_2 = \frac{V_n^2}{100 }$

and not $R_1=\frac{V_1^2}{50}$ ?

12. Oct 14, 2015

### CWatters

The problem statement effectively describes two situations. Resistors tested on their own and then used in a circuit.

The problem statement says (implies) that when each resistor is tested on it's own it dissipates the rated power Pn at the rated voltage Vn. That gives you a way to write an equation for the resistance that you can later use to solve the other circuit..

Pn = Vn/Rn
and
Rn = Vn/Pn

It also says that the rated voltage Vn is the same for both resistors so no need to replace the "n" in Vn with a 1 or 2 because when tested on their own V1=V2=Vn.

You can't use..

R1 =V12/50

in the circuit with both resistors because the power dissipated in R1 (R2) is no longer 50W (100W).

13. Oct 15, 2015

### crom1

That makes sense. Thanks CWatters