Nuts and Volts - February 2018

0.9A+0.36A=1.26A. So, indeed current flows through both resistors, but 2. 5 times more flows through the smaller 10W resistor than the larger 25W resistor.

I like this result as it emphasizes how current and resistance are inverses as per Ohm’s Law. It gives us a mental grasp on the meaning of this sacred law: At a given voltage, more current will flow through paths of smaller resistance; this provides our first hint about how junctions behave. So, perhaps our adage could be changed to “current follows the path of least resistance, but in inverse Albert Einstein was famous for doing “gedanken” experiments, or experiments of pure thought. Let’s do one here to further think about junctions.

Suppose we put a very low resistance (let’s say 0W) jumper wire across the 15W resistor, and add the 50W resistor as shown in Figure 4. What would happen to the current now?

Well, now we have a three-fold junction at A. Current flowing out of the battery travels through the 50W on its way to the junction, where it will flow out, and possibly into 1) the 10W; 2) the 25W; and/or 3) the jumper wire. The current that flows into the junction will “see” all three paths. Which will it choose?

In this extreme gedanken case (involving a 0W wire), all current will flow down the 0W path. It’ll behave as if the 10W and 25W weren’t even present, as absolutely no current will flow through them, for why would current “choose” to flow through any resistance at all when it can flow through a 0W path?

The circuit will behave like the inset circuit in Figure 2, and the battery will supply I=9V/50W=0.18A to it. We added the 50W resistor to avoid (even in a thought experiment) shorting out the power supply, which is yet another experience with Ohm’s Law. Without the 50W resistor, the jumper wire would present 0W across the 9V supply, and the current demand would

¥, blown fuses, overheating components, melting wires, If the 0W resistance was increased — even by a minute amount — we’d then start to see the current distribute itself along all three paths. Neat, huh? Let’s get back to the circuits with the light bulbs and LEDs now.

The Light Bulbs

With the light bulbs of Figure 1, we’ll start with their labels: a 1W and a 2.5W. The Along with Ohm’s Law, there’s a power equation in electronics, commonly stated as P=IV or P=I2R. Note again more mathematical combinations of I, V, and R (but different than Ohm’s Law). The P here is “power” and is in watts, given that I is in amperes and V is in volts. We focus on power here because that’s what light is: some amount of energy-per-second (or power) flowing away from the bulb.

So, on the one hand, since the bulbs in Figure 1 are in parallel (or at the same voltage), P=IV means the more current, the brighter the bulb since P is proportional to I at a given V. This means more current should flow through the 2.5W bulb as it leaves the junction, since we expect it to be brighter.

On the other hand, using P=I2R, we see a stronger (squared) dependence on current, but now there’s an R in there, so does a higher resistance lead to move power at a given current? However, a higher resistance would allow less current flow (I and R are inverses). What gives?

We know that light bulbs work when a resistive filament inside of them gets so hot that it glows. So, let’s start by putting an ohmmeter across the leads of a bulb to measure the resistance of its filament. The 1W bulb gives about 5.5W, and the 2.5W bulb about 3.9W. So, the filament in the higher wattage bulb has less resistance.