Resistance???

Resistors can't do that (well, they can, but possibly not in a useful way).

What are you actually trying to do?

 

edit: I really should refresh the page before I spend so much time repeating almost exactly what BobK had typed several hours previously...
edit2: Actually, I did add a bit more.

All true..

Well it would have been if the advice were good (and it's not).

For powering a bulb you could use a simple series resistor. Beware that for a 5V globe running from 12V, the resistor will dissipate more power than the bulb. Assuming the bulb is a low power one this may not be a problem.

If you know the bulb's power or current requirements you can calculate the resistor you'd need.

Le's assume it's 1W.

The current is determined by P = V * I, so I = P/V = 1/5 = 0.2A

If you already know the current, you could skip that step.

Given a current of 0.2A, a bulb voltage of 5V and a power supply voltage of 12V, you want a resistor which will drop (12-5) = 7V at 0.2A.

Using V = I * R and rearranging to give us R = V/I, we calculate the resistor as 7/0.2 = 35 ohms.

You can't get a 35 ohm resistor, but you could use the next highest easily available value, 39 ohms.

Now we need to calculate the power dissipated in this resistor. This is given by P = I^2 * R (you can derive that from the two formulae given above). This si 0.2 * 0.2 * 39 = 1.56W

Because you want to leave some margin for safety (and so you don't burn your finger if you touch the resistor), you would select one with a power rating of at least 2W, but probably higher would be better, even a 5W one would not be too far over the top.

So, get a 39 ohm 5W resistor.

But remember, you'll need to repeat these calculations for the actual power or current requirement of your bulb.

Another option is to use a 5V voltage regulator. This is a little bit more expensive, but will allow you to run a load that varies from nothing to about 1A with no problems with changing voltage. This will require a 7805 regulator, preferably some small capacitors and almost certainly a heatsink.

If you really only need to run a single bulb, a resistor may be a viable alternative.

 

Essentially, your answer would have been correct if you had considered the bulb as one of the resistors in the voltage divider.

This works because the load (the bulb) can be considered a resistor at its operating point.

To use a voltage divider is to place a resistor in parallel with the bulb. This is rarely a good idea because the best it's going to do is waste power.

 

I don't understand your question.

 

I just got a glimpse of your post before you edited it. All I can say is, good luck. (Because you're going to need it)

If you have a load that requires 2A at 2V, you simply *can't* give it 2V at 0.5A. At 2V it will draw 2A, and at 0.5A it will drop a lower voltage (0.5V -- assuming it is a resistive load).

My method *specifically* uses ohms law. The voltage divider approach mis-uses it.

 

Edit, oh, and by the way 0.07k is not 7 ohms, it is 70 ohms.

Make your life easier by doing all calculations in Volts, Ohms, Amps and Watts. Do the conversion later.

 

Young -- tell me, what resistor values did you use for your voltage divider and how did you model the lamp in the simulator?

I think it's clear that you simply don't realise that once you connect any load to the voltage divider the voltage will fall (and that was essentially the point of the first response I made in this thread).

komalbarun is a false prophet.