Electronic – Light bulb and current limiting resistor

resistors

I am currently interning at a science educational center and for my project, I am creating a simple and educational circuit using batteries, resistors, and light bulbs. The purpose of this circuit is to educate elementary school kids on the concepts of series and parallel circuit. By flipping the switch, they can distinguish the the difference between resistors connected in series and parallel by comparing the brightness.

All 4 resistor values are 1K with 4.5V as input. The specification of the light bulbs are 1.5V @ 0.3A. I have tested all the light bulbs (without resistors) and they are all fully functional. The schematics is shown below.
schematics

However, the bulbs that are connected to the resistor do not light up. When I used the multimeter measuring from B to point A, it reads 4.5V and when I tried to measure from point A with respect to the ground, it reads 0V.

I know this violates KVL as Vf = VR1 + VR2 + Vb. I cannot figure out why I am getting 4.5V drop through the 2 resistors instead of the 2 resistors plus the bulb. I also tried to up the voltage by using a 9V battery, but the results came back the same. Could it be because the light bulbs are not compatible with resistors?

Thanks in advance.

(Note: My supervisor prefers light bulb over LED even though I tried to convince him to use LED…)

Best Answer

the light bulbs are 1.5V @ 0.3A

How do you expect to get 300 mA thru 1 kΩ resistor with only 4.5 V applied? Even with the full 4.5 V across a 1 kΩ resistor, you only get 4.5 mA flowing thru it. With two such resistors in parallel, you get 4.5 mA thru each, so 9 mA total. That's not going to light up a bulb that requires 300 mA for normal brightness. And, this isn't even accounting for the 1.5 V drop across the bulb when it is lit.

One solution is to use lower resistances. Using Ohm's law, we can calculate what they should be. You have 4.5 V available, and the bulb will drop 1.5 V of that when fully lit. That leaves 3.0 V across the resistor. With 3.0 V across it, the resistor should pass 300 mA. (3.0 V)/(300 mA) = 10 Ω. That's the total series resistance for fully lighting the bulb. Since you want that to happen with two resistors in parallel, they should each be 20 Ω.

Also consider the power the resistors will dissipate. Let's be safe and say they need to survive the full 4.5 V applied to them, because stuff inevitably happens. (4.5 V)2/(20 Ω) = 1.0 W. Therefore get "2 W" resistors. Those are also larger and easier to see, so better for demonstration purposes anyway.