Lowering current with 3 100 Ohm-resistors

Oskar has purchased a station building to his model railroad. The lighting in the house consists of 10 light bulbs connected in series, marked 2.4V/0.1A. Unfortunately the lamps often brakes so Oskar decides to change to orange-colored LEDs.

Oskar buys 10 orange colored LEDs and 3 resistors, each with the resistance 100 Ω. The diodes must be connected in series with a current of 20 mA and voltage fall of 2.1 V over each diode, according to the manufacturer's recommendation in the included datasheet.

Help Oskar to design the circuit.

Click to expand...

In the first case 10 lamps with 2.4V/0.1A gives 24 Ohm per lamp, hence are all lamps equivalent to a 240 Ohm resistor. With a current of 0.1 A this gives a voltage of 24 V.

In the second case there are 10 LEDs with 2.1V/20mA which gives 105 Ohm per lamp. This gives an equivalent resistance of 1050 Ohm. Using the same voltage as in the first case, the current through the lamps will be 24V/1050 Ohm = 0.023 A, which is a bit more than the required 0.02 A.

I understand the resistors must be put in the circuit to make the current go down to the required level, but how can we determine how the resistors should be connected? Is one supposed to just try and see if it works, or is there a more systematic method to find out?

This sounds like a hw problem. How ever here is the solution in steps.

Originally there are 10 bulbs @ 2.4v = 24 volts.
The first step is realizing 10 LEDs @ 2.1 is 21 volts.
We specify 20 mA is the desired operating current.
24 - 21 = 3 volts.
3v/ 20mA = 150 Ohms.
we have 3 100 Ohm resistors. to get 150, we connect 2 in parallel which are connected in series with the remaining resistor.
2 R in parallel are 1/2 the R value so 2 100 ohms in parallel become 50. Add the third R in series and we have 150 ohms.
This is an exercise in realizing common R properties. Incidentally, 2 V will be across the series R while 1V is across the parallel Rs.
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Thanks, I understand in this case it's relatively easy to just "recognize" how it should be, as we have 3 100 Ohm-resistors and we want them to be 150 Ohms. But if we are to generalize this, let's say that:

We have x (number of) resistors with resistance y Ohm.
We want them to be connected in a way so that they will together be z Ohms in total.

Is there then a systematic way to find out how to connect them without trying every single combination?

Teszla said:

Thanks, I understand in this case it's relatively easy to just "recognize" how it should be, as we have 3 100 Ohm-resistors and we want them to be 150 Ohms. But if we are to generalize this, let's say that:

We have x (number of) resistors with resistance y Ohm.
We want them to be connected in a way so that they will together be z Ohms in total.

Is there then a systematic way to find out how to connect them without trying every single combination?

Click to expand...

This is an example of applying what you know w/o the rigor of a formula. there are similar exercises like you mentioned. the primary idea is to realize that if the total R is more than a single R then you will need a series connection, if it is less a parallel of some type.

I remember a problem from DC circuits where you have an R in series with another R while the second R has in parallel a Series R with another R with a parallel ... ad infinitum. The trick is to calculate the impedance of the total circuit.

A ---R---|---R---|---R---|---R---
| | |
| | | ...
R R R
| | |
B -------------------------------

Now Assume a value of R=10, calculate the resistance between A and B. There are 2 ways to do this, one is manually calculate until you get a convergence (Brute force), the other is to utilize a property of math and determine it. You might need some knowledge of convergent series.

The problem is NOT to find the value but the method. Good Luck should you choose to try.