Maximum power dissipation for resistors

In summary: Each power resistor should dissipate nearly 500W (full wattage) connected in series and/or parallel. The supply line is 415ac. The delta-wye combination can also be mixed in order to connect the...In summary, the problem is that you need to connect each power resistor with dissipation rating of 500W in series or parallel in order to dissipate the maximum power.
  • #1
jackferz
13
0
there are 3 resistors of, different values. 470o,22o and 100o. but each one is 500W
I want to connect them in series/parallel in order to dissipate the maximum power 500W.

I have connected 4,22o in series and 2, 100o in series, 1, 470 in parallel. but the dissipation is not maximum.
the supply is 415

i would like to know if its possible to get max power dissipation
 
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  • #2
jackferz said:
there are 3 resistors of, different values. 470o,22o and 100o. but each one is 500W
I want to connect them in series/parallel in order to dissipate the maximum power 500W.

I have connected 4,22o in series and 2, 100o in series, 1, 470 in parallel. but the dissipation is not maximum.
the supply is 415

i would like to know if its possible to get max power dissipation

Welcome to the PF.

You only have one each of 22, 100 and 470 Ohm resistors? And you want to maximize the power dissipation with no single component exceeding 500W, or you want the total dissipation of the sum to equal 500W?
 
  • #3
i have each resistor in a large quantity. here is the circuit
 

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  • #4
want to maximize with no single component exceeding 500W. each is rated for 500W
 
  • #5
jackferz said:
i have each resistor in a large quantity. here is the circuit

So what *exactly* are you trying to do? Do you want the total power dissipated to be 500W?
 
  • #6
No, total power will be the sum of the all the dissipated power for each resitor
 
  • #7
jackferz said:
want to maximize with no single component exceeding 500W. each is rated for 500W

jackferz said:
No, total power will be the sum of the all the dissipated power for each resitor

The maximum power dissipated will depend on how many resistors you have. If you have an unlimited number of resistors, you can dissipate infinite power.

Please post the *exact* text of the question.
 
  • #8
ok sorry for the confusion

What i have in the attached circuit.

I will have maximum power for each as followed based on P=I^2*R:

22ohm: 489W (that would be multiplied by 4 since 4, 22ohm are used)
100ohm: 430W (that would be multiplied by 2)
470ohm: 366 (that would be single)

Now if the circuit is connected in such a way that each resistor is connected to dissipated at maximum rating 500W. the total would be the sum of all

As long as each is dissipating close to 500W. the sum of each would be the total
 
  • #9
Did you get my question this time

lets make it simple. i don't care what's the total output power going to be when all the resistors are dissipating 500W each.

for the time i just want each to dissipate 500W (or close).

With the current circuit its way to less than 500w for resistors 100, 470
 
  • #10
jackferz said:
Did you get my question this time

No, sorry.
jackferz said:
Lets make it simple. i don't care what's the total output power going to be when all the resistors are dissipating 500W each.

for the time i just want each to dissipate 500W (or close).

With the current circuit its way to less than 500w for resistors 100, 470

Try this -- what voltage do you need across each of those 3 resistor values to get to 500W? Does that help you start to figure out how you could combine them to always have close to that voltage across them?
 
  • #11
using P=V^2/R,the voltages across each would be

22ohm: 104.9V
100ohm:223.6V
470ohm: 484V

any idea how to combine resistors to get these voltages across it
 
  • #12
jackferz said:
any idea how to combine resistors to get these voltages across it

That's your task, right?
 
  • #13
thats why i am here after not been able to figure out

I don't see of combinations since P=I^2*R= V^2/R

which would be the same as i was trying to achieve current.
 
  • #14
jackferz said:
using P=V^2/R,the voltages across each would be

22ohm: 104.9V
100ohm:223.6V
470ohm: 484V

any idea how to combine resistors to get these voltages across it

jackferz said:
thats why i am here after not been able to figure out

I don't see of combinations since P=I^2*R= V^2/R

which would be the same as i was trying to achieve current.

Can you please show us the original problem statement, word for word? If it's in a language other than English, show us the original and then please give us a word-for-word translation.

The problem makes no sense to me as stated.
 
  • #15
if i were to combine then in order to have required voltages across them, the I^2 current is either goin to burn the resistor or not provide sufficient power.
 
  • #16
Problem: Construct a 3-ph resistor load bank provided the following resistors:

22ohm:500W (Quantity : 15)
470ohm:500W (Quantity:15)
100ohm: 500W (Quantity: 15)

Each power resistor should dissipate nearly 500W (full wattage) connected in series and/or parallel. The supply line is 415ac. The delta-wye combination can also be mixed in order to connect the resistors.
 
  • #17
jackferz said:
Problem: Construct a 3-ph resistor load bank provided the following resistors:

22ohm:500W (Quantity : 15)
470ohm:500W (Quantity:15)
100ohm: 500W (Quantity: 15)

Each power resistor should dissipate nearly 500W (full wattage) connected in series and/or parallel. The supply line is 415ac. The delta-wye combination can also be mixed in order to connect the resistors.

Whew. That's a very different problem from the one you've been trying to describe so far. This is a 3-phase voltage supply, and you have a finite number of the resistors. Check.

Okay, so what do the voltages look like in a 3-phase "415Vrms" system? What is a Delta-Wye Combination?
 
  • #18
please see the attached print

I don't know if this information should have make difference in understanding the situation.
Since even with the 3phase balanced load the current flowing through the resistors are same. This is just delta configuration.

it does not seem to help anyway combining wye connection in that, since Phase voltage would be 240V

Do u understand the taks this time
 

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  • #19
What is the phase-to-neutral voltage for a 415V 3 phase system? What is the phase-to-phase voltage? Are they the same or different?
 
  • #20
As you know

in delta configuration

Phase voltage= Line voltage= So Phase-Phase is 415

If i were to connect it in Wye
Thenk phase to neutral votlage would be 240V
 
  • #21
jackferz said:
As you know

in delta configuration

Phase voltage= Line voltage= So Phase-Phase is 415

If i were to connect it in Wye
Thenk phase to neutral votlage would be 240V

So you have two different potentials that you can play with. Is there any reason why you can't use both forms of connection at the same time (with different groups of resistors, of course)?
 
  • #22
yah you are right, its two different potential either 415 or 240.

But it does not seem to help.

I think the maximum power i can take out for each resistor is as follows (following my previous circuit)

for 22R: 498W
for 100: 430W
for 470: 350W

there is no way i could have combination of series parallel if i need to increase more wattage.
Isnt it
 

Related to Maximum power dissipation for resistors

1. What is maximum power dissipation for a resistor?

Maximum power dissipation for a resistor refers to the amount of power that a resistor can handle without getting damaged. It is the maximum amount of heat that a resistor can dissipate without exceeding its maximum operating temperature.

2. How is maximum power dissipation calculated for a resistor?

The maximum power dissipation for a resistor is calculated by multiplying the square of the maximum current that can flow through the resistor with the resistance value of the resistor. This can be represented as P = I^2*R, where P is the power dissipation, I is the current, and R is the resistance.

3. Why is it important to consider maximum power dissipation for resistors?

It is important to consider maximum power dissipation for resistors because exceeding the maximum power rating can lead to overheating and damage to the resistor. This can result in inaccurate readings or even cause a circuit to fail.

4. Can a resistor handle more power than its maximum power rating?

No, a resistor should not be operated at a power level higher than its maximum power rating. Doing so can cause the resistor to overheat and potentially fail, leading to circuit malfunction or damage.

5. How can maximum power dissipation be reduced for a resistor?

To reduce maximum power dissipation for a resistor, you can either decrease the current flowing through the resistor or increase the resistance value. This can be achieved by using a larger resistor or by using multiple resistors in series or parallel to achieve the desired resistance value.

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