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Old 10-01-2010, 12:28 AM
Brian Carlton Brian Carlton is offline
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Join Date: May 2002
Location: Blue Point, NY
Posts: 25,390
Quote:
Originally Posted by Craig View Post
I'll give it a shot. We are talking about voltage (V), current (I) and wattage (W); the governing equation is:

I * V = W

Therefore, to determine the current used by a 50W, 12V bulb:

I = W/V
I = 50/12
I = 4.2 amps

The resistance through that bulb is:

R = V/I
R = 12/4.2
R = 2.9 ohms.

Now if we put that bulb in a 19.7 V lamp:

I = V/R
I = 19.7/2.9
I = 6.8 amps

So, you are putting 6.8 amps through a bulb that was designed for 4.2 amps; and it blew. There is no reason to think the correct bulb will blow.

Does that help?
Not bad for a guy that's not an EE............
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