For the parallel connected LED and lamp, each has the entire battery voltage across.

When series connected, the voltage across each must *sum* to the battery voltage.

Without any more information than is given, the most likely answer is that the voltage across the lamp, which must equal the battery voltage *minus* the voltage across the LED, is insufficient to produce visible light.

While typing this answer, I see that you've added some pictures. It appears that the total battery voltage is about 3V. Given that many LEDs have a forward voltage in excess of 2V, this leaves less than 1V across the bulb.

Do you have a voltmeter with your kit? If so, measure the voltage across the lamp for the series connection.

Ok, let's take this step-by-step up Wittgenstein's ladder.

Step 1:

- Current is lazy and always takes the path of least resistance.

Given the circuit:

^{simulate this circuit – Schematic created using CircuitLab}

For the current to get from point A to point B it's going to go straight down the simple wire rather than take the more difficult route through the LED. So no current flows through the LED, it just goes straight past.

Step 2:

At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node

Ok, but on the circuit above point A and point B are connected directly together, so they are in effect the same point. The circuit is basically the same as:

^{simulate this circuit}

(imagine the little link bit isn't there - the editor won't let you do diagonal lines).

The current I that flows in must equal the current I that flows out of point A. So if all that's going in is going straight out, there's none left to go up to the LED.

Step 3:

No wire has absolute zero resistance. The same with breadboards. So the actual circuit is more like this:

^{simulate this circuit}

Ok, so we have an fixed voltage \$V_{CC}\$. Let's say this is, for simplicity. 5V. The LED has a fixed forward voltage drop. Let's assume for the sake of argument that it's 2V.

Ok. Let's take the LED out of the circuit initially and just work out the voltages dropped across the resistors R1, R4 and R5.

The total resistance for that section will be 100.002Ω (simply add them together). So the current through them would be \$\frac{5}{100.002} = 49.999mA\$.

Therefore the voltage dropped across R4 would be \$0.049999 \times 0.001 = 49.999{\mu}V\$.

Now if you attach the LED across that resistor it's only going to be getting 49.999µV, which is considerably less than the required forward voltage needed to turn it on. So it won't be conducting as it's not on, so the current through the resistors R2 and R3 will be zero.

Now there are more potential steps in Wittgenstein's ladder, but from here we're getting into the realms of subatomic physics, and even quantum theory, so we'll leave it there for now.

## Best Answer

Each photon emitted from an LED is the product of an electron-hole pair recombination. Charge carriers (electrons/holes) in a semiconductor have an energy distribution which is a function of dopant concentration, density of states of the semiconductor and the Fermi distribution. At elevated temperature the Fermi distribution is stretched out and results in a wider range of electron and hole energies. These are all still concentrated near the bandgap, but increases the probability of a slightly more energetic electron combining with a slightly more energetic hole to emit a photon of higher energy (shorter wavelength). At higher temperatures there are also more phonons available that enable transitions that would otherwise be forbidden. This also broadens the emitted spectrum and allow for slightly sub-bandgap emission.

But temperature also affects the bandgap energy. As the temperature increases, the bandgap energy decreases. This will shift your peak to longer wavelengths at higher temperature.