How to explain these observations on the first load, with two loads in series?

Question

Recently I purchased the Snap Circuits Jr. kit and was working through the different projects with my daughter. We decided to connect two loads in series circuit, powered by the battery source. The first load was a LED light, the second was a fan (a spinning wheel where you place the plastic blades on top).

With the two loads in series, I turned on the switch. I observe the following which I could not explain.

1) Both loads were operating, however, the first load (LED light) was dimmer than if it were the only load in the circuit. Why? I would expect full voltage available before the first load, and hence full voltage drop across that first load.

2) If I introduced physical resistance to the movement of the second load (the fan) with my finger, the first load (LED light) would become brighter. Inversely, if I removed any physical resistance (finger, the fan blade) such that the wheel spun freely, the first load would become brighter. Why?

In summary, given a series circuit I don't understand why the second load would impact the first? Please help this uninformed dad so my daughter thinks I know everything again!

Answer 1

Please note that when circuit is connected in series connection, same amount of current flows through all the loads connected in series. However, voltage drop across each load is proportional to the resistance of that particular component. And the sum of voltage drop across each component equals voltage provided by the source.

please refer https://en.wikipedia.org/wiki/Series_and_parallel_circuits.

Case 1: When fan is connected lad glows dimmer.

This is because when the additional load is connected the voltage provided by the battery is now shared by both the loads. (In this case, i.e., when motor is used, current restriction is because of back emf - finally voltage across led equals current multiplied by resistance of led)

Case 2: When you put physical resistance to the movement of the blades.

As explained below, back emf and thus resistance to current decreases when you slowdown the fan. This will increase current in the circuit and led will glow brighter.

When a motor is turning, that rotation generates a voltage, a 'back EMF', that acts against the flow of current. It is this voltage, not the resistance of the coils, that restricts the amount of power the motor draws. And as this is an impedance, it doesn't generate heat. The power - the current in the motor pushing against this voltage - is what turns the motor.

When the rotor is locked, there is no back EMF to impede the flow of current through the motor. All the electricity flowing through the motor is converted to heat by the resistance of the windings. This quickly overheats the wiring, melting insulation, creating shorts, reducing the resistance and further increasing the current, until some wiring melts and blows.

https://www.reddit.com/r/askscience/comments/67tp73/if_an_electric_motor_is_supplied_power_but/