Sorry once again Budd, but it really doesn't work that way. While it is
true that a bulb converts electrical energy into heat and light, it has
nothing at all to do with the conversation. I could go into a long winded
explanation but you would just ignore it so I'll just use some simple
equations. I take it that you are familiar with Ohm's Law, where voltage
represented by E in the equation E = I * R with "I" being current and "R"
being resistance. Power can also be calculated as P = E *
I where once
again "E" being voltage and "I" being current. Using these equations, the
current flow in a circuit "I" can be calculated if the voltage "E" and
resistance "R" are known by rewriting the formula as I = E / R. I probably
shouldn't need to go any further but since it is you, I will. As you should
know, if you increase the size of the denominator in a fraction, its value
goes DOWN, not up. For an example, if our battery is putting out 12V and we
have 6 ohms of resistance in our imaginary circuit (lets say the bulb),
using Ohms calculation of I = E /
R we have I = 12 / 6 which equals 2 Amps.
Now we will make our connections really dirty and for the sake of keeping
the math simple for you, they are so dirty they add another 6 ohms and
effectively double the resistance to our circuit. Now our calculation comes
out to I = 12 /
12 which is the original bulb plus our really dirty
connection and now the current flow is just 1 amp. Since the capacity of a
battery is measured in amp hours, it should be obvious to you that the added
resistance will allow the battery to go twice as long.
As for Power or "P", we can calculate that as well. In our first example,
the battery put out 12 volts and 2 amps were passing thru our circuit so P E * I or P = 12 *
2 which means our bulb was providing 24 watts of light.
With our second example that included you dirty connections, P now equals 12
* 1 where the current was effectively cut in half due to the added
resistance which means our bulb is now only providing 12 watts of light so
as I said in the previous post, resistance in his connections will extend
the life of his battery but at a reduced light output. Of course, these
simple calculations make assumptions about the characteristics of a light
bulb that are not completely accurate but that has nothing to do with the
point being made and that is adding resistance to a series circuit with a
fixed voltage reduces the current draw and power of the circuit.
If at first you don't succeed, you're not cut out for skydiving
"Budd Cochran" < firstname.lastname@example.org> wrote in message
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