wattage question

[lokeycmos]

say i have 2 identical resistors, both value and wattage rating. if i put them in parallel, i get half the resistance but double the power. if i put the same resistors in series i get double the resistance, but what do i get for power rating?

now, say i have two resistors of different values and watt ratings. what would happen in this same senario?

 

[ericgibbs]

say i have 2 identical resistors, both value and wattage rating. if i put them in parallel, i get half the resistance but double the power. if i put the same resistors in series i get double the resistance, but what do i get for power rating?

now, say i have two resistors of different values and watt ratings. what would happen in this same senario?

hi,
W = V *I, so double the resistance halves the current, so halves the W.

 

[Ian Rogers]

The power is shared between them!! You can't deviate from ohms law.. If they are in series they will consume the same as they are taking exactly the same current.

Eric got there first ( as usual )

 

[crutschow]

If by "rating" you mean the maximum power dissipation, then two identical resistors will have double the power rating of one, both in series and in parallel.

If the resistor are different values and ratings then you have to calculate each individual case to determine the resistor that dissipates the most power. The one that reaches its power limit first is the limit for the circuit.

 

[Ratchit]

lokeycmos,

say i have 2 identical resistors, both value and wattage rating. if i put them in parallel, i get half the resistance but double the power.

Are you driving the circuit with a constant voltage, or a constant current?

...but what do i get for power rating?

The power ratings of the resistors do not change. The dissipation depends on the topology of the circuit.

Ratch

 

[MrAl]

Hi,


For two resistors in series we know that the total resistance is:
RT=R1+R2

and that the total power rating when both resistors are used to their fullest extent (not necessarily full power) is:
PT=P1+P2

and so we get a nice equality:
P1/P2=R1/R2

This tells us that to use the two resistors to their fullest extent (to create a higher power resistor) the ratio of their power ratings must equal the ratio of their resistances. Since the current goes down as the resistance increases, this means either resistor value can be larger than that in this equation but not smaller, and it almost goes without saying that the power rating of either or both resistors can be larger than that shown in this equation.

This means that for example we can calculate the required power rating of a second resistor to be placed in series with the first knowing the resistor's value (which we should know because we have to know that the total resistance is going to be) using:
P2=(PT*R2)/(R2+R1)
where
PT is the total power desired,
R1 is the first resistor value,
R2 is the second resistor value,
P2 is the required minimum power rating of R2.

 

[lokeycmos]

so say i needed a 450.45Mohm resistor rated for 22watts and i wanted to use multiple series resistors to equal those values and i used 10 resistors. each resistor would be 45.045Mohms. does that mean each resistors wattage rating would be 22/10 = 2.2 watts?

 

[ericgibbs]

so say i needed a 450.45Mohm resistor rated for 22watts and i wanted to use multiple series resistors to equal those values and i used 10 resistors. each resistor would be 45.045Mohms. does that mean each resistors wattage rating would be 22/10 = 2.2 watts?

hi,
Dont confuse the resistors Wattage Rating, which is normally specified in the datasheet, with actual Wattage Dissipation when in use due to the current flowing thru it.

Why are you working with such high value resistors.???

 

[lokeycmos]

hi,
Dont confuse the resistors Wattage Rating, which is normally specified in the datasheet, with actual Wattage Dissipation when in use due to the current flowing thru it.

Why are you working with such high value resistors.???

Making a high voltage probe for my DMM 100:1 ratio. i realize i must use several resistors in series. i did the math last night and the 450.45Mohm resistor must dissipate 21.756 watts. just trying to figure out what the individual wattages must be when i use multiple resistors in series. im not using anywhere near that much voltage, just using it as a maximum value so i have overhead.

heres i pic of circuit

 

[ericgibbs]

hi,
Whats the input impedance of the DMM.??

Assume its 1megOhm, so for a 100:1 divider you would require a 100megOhm resistor in the top of the divider and a 1megOhm as the bottom resistor, thats IF you wanted to retain the Ohms/Volt of the original DMM input as seen by the high voltage you are wishing to measure.

BUT if the high voltage source output impedance was 'low' and could tolerate the loading of say a 10megohm resistor, it would be possible to make a input divider of 10meg and a ~0.1megOhm resistor.

Assume that your resistors are rated for a maximum working voltage of 250V and that you used 10 of 1megOhm in series to make the 10megOhm, thats 2500V max for the resistor chain.

Assume that you want to measure 2500Vmax, the current would be ~250microAmps.

Total dissipation across the chain will be ~0.625Watts, so each resistor would dissipate 0.0625W, this means the 1megOhm resistors could be rated at 0.125W [ preferred value].

E.

 

[lokeycmos]

will will voltage divider for DC DMM work with AC?

thanks to forum members i have the math to make a 100:1 voltage divider to extend the range of my DC DMM. my question is, will this resistive divider work equally well to measure AC? if it doesnt, how and what do i have to modify to make it accurately measure AC as well as DC. thank you in advance!

 

[JMW]

Recheck your math. As for the power, regardless of whether they are series or parallel they CAN still dissipate rated power. When in series, the voltage will have to be increased to raise the current. Remember to use hi voltage resistors when you do this. Insulate them well (means overate the wattage), connect the leads with the equipment off, power up take the reading, then power down. Making certain you wait several seconds after the meter reads 0 indicating the circuit has discharged.