Quote:
|
Originally Posted by Jim H
P1 = V1^2/R
P2 = V2^2/R
P1/P2 = V1^2/V2^2
(^ is exponent, in this case squared)
Say P1 = 7.0V, P2 = 6.0V
P1/P2 = 7x7/6x6 = 49/36 = 1.36
On 7.0V, the 6.0V lamp will consume 1.36 times its rated power, so the 30W bulb will consume 41W. It will be brighter, by about a linear ratio of 1:1.36.
Best Regards,
Jim
|
but not quite correct! The formula is :
Pr = E1^2/E2^2 x R2/R1
where
Pr = power ratio between two dissimilar bulbs,
R2 = resistance of lower voltage bulb,
R1 = resistance of higher voltage bulb.
You can't ignore the resistance of either bulb since, as the formula shows, the resistance ratio will further worsen the power ratio because the lower rated bulb will have higher resistance. This will increase the power ratio between two dissimilar bulbs.
So for your example of "7V" and "6V" bulbs, the resistance ratio would make the ratio equal to
1.36 x R2/R1. Thus the 'final' ratio will be larger than "1.36" by the ratio of the two bulb's resistance. The only time you can ignore the resistance ratio is when the two bulbs are equal but this a 'trivial' case.
Bottom line: the stress on the two
dissimilar bulbs is greater than just the ratio of operating voltages. You need to include the resistance. But in any case, the consequence is a very short, but very bright, life.