Does a 12V light = a 12 V voltage drop? Help please

[pete]

Hi,

Suppose a 12V battery is hooked up to a lamp that requires 12 volts.

Theory tells me:
In a series circuit, before the lamp, the voltage is 12v and after the lamp,
the voltage is 0v.
Is that true? Is that called voltage drop?

Let's say I hooked up a voltmeter to this circuit. What would it read and how
would one interpret it?

Any questions you could tackle would be helpful. Thanks!!!

Pete

[Lord Garth]

Voltage is measured across a circuit...since there is no other component
in your example, there is nothing else to measure 'after' the lamp.

You have described measuring the battery only.

pete <pch...@canada.com> wrote in message
news:wGHh4.4862$mK.3...@brie.direct.ca...

[Peter Russell]

Why don't you just try it?
--
----

PeteR

[Gottfried Pauler]

Well, voltage is not something that is here or not - it is potential difference
between two points in a circuit. Therefore when you say "There is a voltage of
???V.", you also have to say between which two points.

Your example:
When you measure the voltage between the connectors of the lamp you measure the
voltage-drop of the lamp - and that is 12V in your example. If you have a very
strong lamp - which means that much current flows - and you have rather small
wires, then the resistance of the wires from the battery to the lamp will also
show up a considerable voltage drop. Then the fact may be that the voltage drop by
the 2 wires is 0.5Volt each and 11Volt at the lamp. The sum of all voltage drops
has to be the source voltage. The rule is: The highes resistor drops the highes
voltage! And also use Ohms-Law: Resistance=Voltage/Current

Regards,
Gottfried

pete wrote:

> Hi,
>
> Suppose a 12V battery is hooked up to a lamp that requires 12 volts.
>
> Theory tells me:
> In a series circuit, before the lamp, the voltage is 12v and after the lamp,
> the voltage is 0v.
> Is that true? Is that called voltage drop?
>

> Let's say I hooked up a voltmeter to this circuit. What would it read and how
> would one interpret it?
>

[Mark Kinsler]

pete <pch...@canada.com> wrote:
>Suppose a 12V battery is hooked up to a lamp that requires 12 volts.
>
>Theory tells me:
>In a series circuit, before the lamp, the voltage is 12v and after the lamp,
>the voltage is 0v.
>Is that true? Is that called voltage drop?
>
>Let's say I hooked up a voltmeter to this circuit. What would it read and how
>would one interpret it?

This circuit isn't as simple as it looks. First off, it's a unique
situation in that the two circuit elements are connected both in series
and in parallel. The voltages across two parallel circuit elements are,
by definition, identical. The current through the battery and through the
bulb are identical as well, because they're connected in series.

The other problem is that your readings on a voltmeter would be somewhat
different from what you'd calculate. The calculated value would, of
course, be 12v across both battery and lamp. This assumes that the wires
have zero resistance and that the voltage source is an ideal one.

But while the wires probably don't have enough resistance to worry about,
that voltage source is always a concern. Turns out that every battery or
generator has an internal resistance that's effectively connected in
series with it. That gives your circuit not two, but *three* elements: an
ideal 12v source, the internal resistance of that source, and the bulb.

Thus, if you measure the battery voltage when the battery's not connected
to anything, you'll measure a dandy 12.00 volts, or whatever it's supposed
to be. But when you hook up that lamp, you'll measure maybe 11.93v across
the battery or the lamp. That's because current is flowing through the
battery's internal resistance and lowering the voltage you measure at the
battery's terminals. The difference between 12.00 volts and 11.93 volts
is the voltage drop across that internal resistance. The lamp will drop
the remaining 11.93v and that'll give you the result you'd expect, which
is that the sum of the voltages measured around any loop will be zero.

M Kinsler

--
............................................................................
114 Columbia Ave. Athens, Ohio USA 45701 voice740.594.3737 fax740.592.3059
Home of the "How Things Work" engineering program for adults and kids.
See http://www.frognet.net/~kinsler

[Michael L. Williams]

Very well said! Nothing like a little Kirchhoff keep those voltages
straight!

Also check out:

http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.KVL.html

for a graphical explanation of KVL

GoldChain

Mark Kinsler <kin...@frognet.net> wrote in message
news:BiLh4.17297$pb2.1...@tw11.nn.bcandid.com...

[The Team]

>Why don't you just try it?
>--
>----
>
>PeteR
>

I agree also, get your hands on any multimeter, wire up a bulb and
battery and measure the voltages.

If you don't own one try looking at Maplin Electronics. Personally if
you are learning I wouldn't bother to buy a cheap analogue multimeter
- I ruined 3 when I was learning ! I've had a look at the Maplin
catalogue and they do digital ones from £8 ( you can ruin these also
but not as easily)

Rod

[Phil Robinson]

and of course Mark has only mentioned the resistance part of the battery,
which will have both capacitance and inductance as part of its total
impedance. This means (if we're fussy) that the voltage may change in the
micoseconds after switch-on.