Can a 10W 5ohm resistor take 100V for 3 seconds?

[rg12]

Are the results affected by different ohm of different resistors?
If for example I use a huge resistor with 5ohm vs 10ohms etc...

 

[Solcar]

Are the results affected by different ohm of different resistors?
If for example I use a huge resistor with 5ohm vs 10ohms etc...

A 10 ohm resistor draws less current than a 5 ohm. The battery internal resistance is somewhat constant, however.

I see it is internal resistance rather than infrared that needs to be determined.

A 1000w @ 120v halogen light bulb could be used in the process of determining the internal resistance of the battery. A multimeter that has a 20 ampere scale is useful in this determination, too.

Measure the no load voltage on the battery. That is E. Then hook up the light bulb load and measure loaded voltage, V. Next set the meter on the 20 amps DC scale and measure
current drawn by the load, I. The meter goes in series with the load for that measurement.

Now E, V, and I are known. Calculate the battery's internal resistance



Internal R = Voltage drop inside the battery / I


E - V = Internal voltage drop of the battery

Internal R = (E - V) / I

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An alternate way is to improvise a dummy load so that taking a current reading isn't needed. Measuring the resistance of the dummy load permits the calculation of I instead of having to measure it.

The resistance of a heating device like an infrared heater can be measured. After placing it across the battery leads, measure the load voltage. Then current, I, can be calculated using the ubiquitous formula,

V = I * R, and so, I = V / R. That is the current drawn through the load, and also through the battery.

Changing things around by solving for R gives us R = V / I

And further,

E - V = Internal voltage drop of the battery

So

Internal R = (E - V) / I