Doubts about resistors in series and parallel

The powers of the resistors in parallel are added!!!

And the powers in series???

Are added or lesser power prevail???

Some people say that add, other than the smallest prevails!!!

Which statement is correct???

Sorry for my bad english!!!

10x

To calculate the total consumed power;
You have to add the powers of all resistors or any components in the circuit.

Suppose I have a resistor in a circuit that was 100R/5W, but he burned!
He was correctly sized for the circuit. But there was a natural component wear.
However, I have resistors and 50R/1W 25R/2W ... Making the association in series of these values​​, the power will be 100R/1W or 100R/5W???
So to say that in this case have the 5W!!!

10x

nagkiller said:

... I have resistors and 50R/1W 25R/2W ...
Making the association in series of these values​​, the power will be 100R/1W or 100R/5W???
So to say that in this case have the 5W!!!

Click to expand...

You need 4 * 25R/2W in series, together they can stand (dissipate) 4*2W=8W.

50R + 25R + 25R in series doesn't work in your case, because - say from a total power dissipation of 6W - the 50R/1W would get 3W - too much for it.

The power, you are indicating with 100R/5W representation is the max power (The power the device will stand without burned).
But you should calculate power from current (From the formula i=V/R and P=i^2*R)
If the power from this formula is higher than 5W (for the 100R/5w) than the resistor will burn.

And 100R/5w doeesnt mean the resistor will drawn 5w in all stuation.

This subject is getting interesting!!!

How do I calculate the power of this circuit (combination of resistors)???



3W, 6W or 12W???

nagkiller said:

This subject is getting interesting!!!

How do I calculate the power of this circuit (combination of resistors)???

View attachment 104966

3W, 6W or 12W???

Click to expand...

12w of course.

V across each resistor can be sqrt(3 x 100).
Hence across the whole thing it is 2 x sqrt (300)
And we know total R is 100

Hence total power is V^2/R = 4 x 300 / 100 = 12

In the example I mentioned, we will forget everything except that the component was a resistor 100R/5W he had wear and tear, then the circuit diagram, I know he is a resistor 100R / 5W, but I have this component on my workbench, I have only resistors 50R/1W and 25R/2W, if I add 1 resistor 50R + 2x25 = 100R. But if the doubt would be the total power dissipation.
If the original component was 5W, I have to insert another 5W or greater.
So I wonder if my association would be ideal?
Or, for example using two resistors 50R/3W, in which case adding, would 6W!!!
But in parallel to achieve the same 100R, I would have to make an association in series of 200R and then insert another association of 200R in parallel with the first, totaling the 100R, but what about power???
Associations in series, the power is added, but in parallel???

- - - Updated - - -

To finish my examples:



What is the total power??? 16Watts???
Only the Power (Watts) in association!!!

Let suppose we have a single resistor R to which we apply Vin volt. The power dissipated will be:

P = Vin²/R

we could also use the current flowing through ther resistor I=Vin/R then

P = R*I² (of course we will have the same value as that calculated by using P = Vin²/R)

Now we want to use two resistor R/2, in series, to form again a resistor of value R. Let's calculate the power dissipated by each resistor.
Since the two resistor have the same value (that is R/2) the voltage across each resistor will be Vin/2. Let's calculate now the power dissipation of each resistor:

P = (Vin/2)²/(R/2)=(Vin)²/(2*R)

So each resistor have to dissipated half the total power. This means, for instance, a resistor 100 ohm, 10 W can be repalce by two resistor in series 50 ohm, 5 W.

If instead we want to use two resistor in parallel 2*R to form a resistor of value R the voltage across each resistor will be the same, but the current will be an half with respect to that flowing in the original resistor of value R.
In detail the current flowing on each resistor will be:

I=Vin/(2*R)

the the power dissipated by each resistor will be:

P=2*R*[Vin/(2*R)]²=(Vin)²/(2*R)

that is the same calculated by two resistor in series. This means, for instance, a resistor 100 ohm, 10 W can be repalce by two resistor in parallel 200 ohm, 5 W.

In your case, since all the resistor have the same value, the maximum power that can be dissipated is the sum of the power of all the resistors (of course the maximum voltage of each resistor have not to be exceeded).

In short: In series or parallel, the Power (Watts) are summed!!!

Thanks!!!

nagkiller said:

In short: In series or parallel, the Power (Watts) are summed!!!

Thanks!!!

Click to expand...

No.

This is only for the special case where all the R's are same.

A counter example : 2 r's in series, 100R/ 10w and 100R/ 1W

Max power will be 2W before the smaller one breaks.

I refer to the previous example!!!
All resistors are of the same power!!!

All 100R / 2W!!! In this case, all power is added!!!

View attachment 104969

My doubt is not the association in series or parallel, but the sum or no power (Watts)!!!

As I said, its added ONLY for your special case/ circuit.

Another example : if even one of your circuits resistors is (say) 1W, then max power suddenly becomes only 8W !!!