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Why does less resistance dissipate more power

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SanjKrish

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I have two resistances which I will call them as loads... and they are purely resistive...
First one is a heavy load( A geyser which has 1000Ω resistance coils to heat up the flowing water)
Second is a very light load( A hand held electric fan of 10Ω )





Intuition says the geyser should be consuming more power as it is a heavy load..:idea::idea:
but calculation shows it otherwise...:shock::shock::shock:

Help me understand:?::?:
 

Resistance is just a measure indicating how easy/hard the current can flow and it does not directly dictate how much power it would consume. Power, in pure physics terms, is how much work the system can do in a unit time, and amount of current and potential would directly determine the power by definition. Of course, you can relate the power to the resistance since these three properties are linked by Ohm's law and it can increase/decrease the amount of current, but by definition, current and potential should determine the power.

A simple analogy. Two idential water towers, and one with wider pipe(low resistivity) down to the ground, and another with narrower pipe(high resistivity) down to the ground and run the water through those pipes.. Which one causes more damages on the ground in a given time ? That's the interpretation of power in physics.
 
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Intuition says the geyser should be consuming more power as it is a heavy load..:idea::idea:
but calculation shows it otherwise...:shock::shock::shock:

Help me understand:?::?:

Heavier load is the one with the lower resistance and the resistance in that case is lower for the fan,
your calculations have to do only with the given data not with the image of the devices.
It is true that in reality the geyser would consume more power but for the data given for calculation your results are correct.

Alex
 

Heavier load is the one with the lower resistance

If it is so, then the copper wires which connect the load and the power supply should dissipate the most power and eventually get melted, as it offers the least resistance in the circuit...
But it doesn't???
 

Load is considered the total resistance seen by the source,
you have 2 circuits with the same source but different loads, in that case the heavier load is the one with the lower total resistance because it forces the source to provide more current.
When you have 2 loads in series then the current provided from the source is the voltage divided with the total resistance and the current is common for both loads,
the voltage on the other hand will be more on the load with the higher resistance (because of the voltage divider formed) and the power dissipation will be grater on that load that has more resistance.
The wire is a heavier load if you connect it as a load not in series with a second load, for example calculate the curent for a wire with resistance of 0.1 ohm.

Alex
 
Yes alexan..,
I guess I am getting it now... I have read tat if two terminals are shorted by a wire... the whole line will get heated up and tats why we go for fuse....

So a 100Watt bulb should have lower resistance that a 60Watt bulb.... Is it right??
 

So a 100Watt bulb should have lower resistance that a 60Watt bulb.... Is it right??

Yes, assuming that both are specified for the same voltage.

You only need two equations to do these calculations,
ohm's law I=V/R (which can also be transformed to V=I*R )

and P=V*I,
this can be also transformed to P= (I*R) *I which is the same as P=I²*R
or you can transform it as P=V*(V/R) which is the same as P=V²/R

using these equations you can calculate the power over a resistor using either V&I or V&R or R&I

You can also calculate the resistance from the power and voltage, P=V²/R can be transformed to R=V²/P,
so in your question above if the bulbs are intended for 220v then

the 100W bulb is 220v²/100w=48400/100=484 ohm
and for the 60W bulb 220v²/60w=48400/60=806.66 ohm

Alex
 
When you actually take the resistance of a 60W light bulb, the resistance measures 20 Ohm. I wonder why?
 

When you actually take the resistance of a 60W light bulb, the resistance measures 20 Ohm. I wonder why?

Because tungsten filament light bulbs resistance is much lower when cold than when at the working temperature

Alex
 

I have two resistances which I will call them as loads... and they are purely resistive...
First one is a heavy load( A geyser which has 1000Ω resistance coils to heat up the flowing water)
Second is a very light load( A hand held electric fan of 10Ω )

Intuition says the geyser should be consuming more power as it is a heavy load.. but calculation shows it otherwise.
Help me understand

This question is just a joke? A normal hand-held electric fan is about. 10 to 50 watts, while a flow boyler consumes 2 to 4 kW. In other words, you are probably reversed the measured resistor values ...

And, the fan is not purely resistive (motor!).
 

Incandescent light bulb - Wikipedia, the free encyclopedia

"Electrical characteristics

Incandescent lamps are nearly pure resistive loads with a power factor of 1. This means the actual power consumed (in watts) and the apparent power (in volt-amperes) are equal. The actual resistance of the filament is temperature-dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms).[33][64] Since incandescent lamps are resistive loads, simple triac dimmers can be used to control brightness. Electrical contacts may carry a "T" rating symbol indicating that they are designed to control circuits with the high inrush current characteristic of tungsten lamps. For a 100-watt, 120-volt general-service lamp, the current stabilizes in about 0.10 seconds, and the lamp reaches 90% of its full brightness after about 0.13 seconds.[65]"

Alex
 
Because tungsten filament light bulbs resistance is much lower when cold than when at the working temperature

Alex

This leads to why lightbulbs typically burn out right when you turn them on. At the moment of turn-on, they are drawing: 220^2 / 20 = 2420 watts! Once the element heats up, it's resistance increases, such that at steady-state it consumes 60/100 watts.
 

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