Tuesday, June 25, 2013

Resistors in parallel, part 2

Again, I refer the reader to the HO:ME blog, which really should be read in parallel (pardon the pun) with this one.  As we learned last week, two blogs in parallel should reduce your resistance to learning...

Today a variation on the theme from last week, in that we will wire two resistors in parallel, that have different resistances.  We spent some time last week looking at a accidental case of this, in which we wired our "triangle circuit" 

And saw that the flow through the wire should be 10^5 higher through the wire than the resistor.  My little brother saw this and informed me that in a circuits class, this would be viewed as "shorting out" the resistor.  The wire would be assumed to have a resistance of zero and the triangle above would simply be erased.  I actually measured the resistance of the wire (0.4 ohms) and although I am not perfectly certain my multimeter is accurate down in those ranges, it's the data I got, so it's the data I'll use. 

So the setup of today involves a 1K and a 2.2 K resistor, like so:

I have drawn these out using Modified Deshazer Notation, which probably does not coincide with standard notation. 

So let's see what these look like.  Let's first demonstrate the baseline current for each resistor:

So if you can't read that, it's 0.74 mAmps through a 2.2 Kohm resistor on the left, and 1.61 mAmps through a 1Kohm resistor on the left. Baseline voltage is about 1.6. 

And then each one in parallel, just measuring off one of situation A) or B) above.

The pic above is measuring across the 2.2 Kohm resistor arm of the parallel resistors.  Note that the measured amps are identical to the initial, single-resistor-only pic above.

Here is the measurement across the 1 Kohm resistor:

Interesting, no?  It appears that the current across either arm of this setup is identical to the measured amps prior to hooking up the parallel resistor.   I confess I had expected to see a change in the current across each resistor as I added a second one in parallel.  I guess this makes sense, as the voltage drop across each of these resistors is identical.  However, the resistance has changed, so let's measure the total current in the circuit:

The components of the current seen on each arm add to (more or less) the current seen across the entire circuit.  Thus, the more (parallel) resistors you add, the more current you get!

In fact, for certain circuits, this may be an easier way of finding the resistance.  As the current will be the same across each resistor, regardless of how many others will be placed in parallel, it may make sense to calculate the sum of those currents, then use Ohm's law to determine the total resistance of the circuit.  Seems like it could be easier than 1/R = 1/R1 + 1/R2, etc.

So, let's check.  The current is inversely related to the resistance, therefore 1/R = (1/1000) + (1/2200), or 687.5 ohms.  Total current we measured at 2.33 mAmps.  V=IR gives us a calculated voltage of 1.6 volts, which is an accurate reflection of our measured volts.

Good learning!

No comments: