how can i calculate the required resistor for a circuit using voltage and the required voltage?

You need the current.

Subtract the two voltages and divide by the current.

for the power required in the resistor multiply the voltage difference by the current.

Be wary, if the current changes, and it probably will, the voltage will change.

For most applications, you should use a 3 terminal regulator.


edit: you asked how you get the current, that depends on what the device is that is the load. More details will help.


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You need a bit more detail to give an exact answer. The most obvious is to use ohms law.

It says V=IR where V is voltage, I is current (in amps) and R is resistance in ohms.

Equally, you can derive I=V / R of R = V / I

Also be aware of wattage and dissipation. This can be done with W=VI - again, W is Watts, and V and I are voltage and amps.

If you have a steady voltage of (say) 20V and want to get 12V (it must be a load which uses a constant current, otherwise things change), but assuming your load needs 350m/A;
- you need to drop 8V across the resistor, leaving 12V for the load.
- the resistor value is R=V / I = 8 / .35 (remember, it has to be amps here), so the answer is 22.8 ohms. This is not a standard value although 22 ohms is, so you would use this.
- the wattage is W=VI = 8 (across the resistor) x .35 = 2.8W

2.8W resistors are non standard so I would use a 5W wirewound (be generous here, use higher ratings, never lower).

All this assumes everything is static and unchanging.

3 terminal regulators, as already suggested are a good idea. They need a slightly higher input voltage than what you want to derive, but not too many volts over - as this is dissipated as heat by the regulator. For non-standard volages, regulators may be made variable using a single resistor and a pot. Without knowing your exact needs, I would be suggesting you research 3 terminal regulators on Google (some common part numbers are LM309 or LM309K for the steel body, 78xx where xx is the voltage, such as 7805, 7809, 7812, etc, LM338K - a real beauty delivering about 10 amps, etc).

Yet another option is to use a Zener diode, but again, depends on circumstance, which you don't give enough of. To give you a decent answer you must expand upon your question.

Good luck.

EDIT: I have just read the first answer paying more attention, and these two answers are identical in fact. He said it in fewer paragraphs - lol.

ANOTHER EDIT: To measure th current, insert an ammeter in series with the circuit drawing the current. Be sure your range setting is higher than the expected current or you may damage the meter. A good way, without interfering with the circuit is to use a clamp ammeter. This has jaws that open and you place it around a conductor to the power supply. It reads the current through induction, and are much harder to blow - and save breaking into the circuit. As a general rule, they work better on high currents than low ones (so far as I know - I have not seen one working in the milliamp range). What type of circuit are you working on? Email me at brett2010@ymail.com if you want more info.

I believe that what you're thinking of is called "voltage divider" equation.

It says that: If you have two resistors (R1 and R2) in series with a voltage source (Vs), that the voltage drop (VR1) across R1 is simply:

VR1 = ( R1 / (R1 + R2) ) x Vs

Then, of course, the voltage drop across R2 is simply the difference between Vs and VR1.  Which, when you add VR1 and VR2 it should add up to Vs.

Current, in terms of ohm's law, is:  I = E / R, where I is current in amps, E is voltage, and of course R is resistance in ohms.

1 amp of current = 1000 miliamps --- a conversion factor
ex. 100 mA of current is 0.1 Amps.

There isn't enough information here to tell you how much current you would need to operate your device.  You will need to know the operating resistance of the device and then plug those values into I = E / R formula.  However, for a 9 volt battery operated device, you can figure an approximate operating current in the neighborhood of 50 to 100 mA.