1.1A through the 1W bulb. Thus,
about 1.3 times more current
will flow through the 2.5W bulb
as it leaves the junction than will
flow into the 1W bulb.
Here, P=IV indeed provides
a clear picture of why the 2.5W
bulb is brighter: its filament has
less resistance, and at a given
voltage, it draws more current
(via Ohm’s Law), and since light
power (P) is proportional to I, the bulb will be brighter.
More current leaving the junction at A will “choose” to
flow through the 2.5W bulb than through the higher
resistance of the 1W bulb.
To test all of this, I fastened
two bayonet-style bulb holders
onto a scrap piece of wood (as
shown in Figure 5) to hold the
1.0W and 2.5W miniature bulbs.
The red clip lead is from the
positive of the power supply,
and the yellow is to ground. The
white, lead, and (soldered) green
wire complete the parallel
configuration. The 2.5W bulb is
clearly brighter. This might all seem obvious at this point,
but there’s a bit more to these light bulbs. Let’s dig
deeper.
Current vs. Voltage
In electronics research, a useful first step in
understanding an electrical device is to acquire “I-V” data
for it: “I” stands for current, and “V” stands for voltage.
Such data means to apply a given voltage across a device,
then measure the current it draws at that voltage. If this is
done for a sequence of voltages — for example, from 0 to
10V in steps of 0.5V — a data set can be obtained, then
graphed.
A simple I-V data acquisition circuit is shown in Figure
6. You’ll need two meters to do this: the one labeled “A”
is an ammeter (or a “voltmeter” set and wired in current
measuring mode); and the one labeled “V” is a voltmeter
(or a second “voltmeter” set and wired in voltage
measuring mode). A device to be characterized is inserted
for the “device” in the figure. The arrow through the
battery symbol means an adjustable voltage supply, and
the resistor R is used to protect the unknown device, in
case it itself has a very low resistance. (I actually omitted it
in the work here.)
If you build this tester, it’s fine to cobble it together
using clip-leads and alligator clips. Just try to keep the
wires neat, and work carefully to insert your meters
properly. I’d build the circuit first and be sure the bulb
glows. Then insert the ammeter, again being sure the bulb
glows and that you see a current reading. Lastly, clip the
voltmeter across the bulb (and be sure it still glows).
In Figure 7, I-V data is shown for a resistor placed in
the device of Figure 6. I turned the voltage on my power
supply all the way down to zero, and recorded zero
current. I then turned the voltage up to 0.5V and read
about 0.0011A, and so on. I typed the data into a two-column spreadsheet and made an XY-scatter graph.
From the graph, I see that the higher voltage placed
across a resistor, the more current it draws. This sounds a
lot like Ohm’s Law, in that I=V/R (I is proportional to V),
and since the data looks linear, I’ll find a “trend line” for it.
When I did so, the slope came out to be 0.00212A/V.
What of this?
32 February 2018
FIGURE 7. Current vs. voltage data acquired for
a resistor inserted for the "device" of Figure 6.
FIGURE 8. Current vs. voltage data acquired for
2.5W and 1.0W light bulbs.
FIGURE 6. By slowly ramping the voltage
of the power supply, this circuit would
allow one to acquire a voltage vs. current
dataset for the "device."
