2006-11-13 13:16:51 · 7 answers · asked by doc m 1 in Science & Mathematics ➔ Engineering
It depends on the current. You want to drop 6 V across the resistor to get the voltage to the load to 6 V. The voltage drop across the resistor is V = R*I. The resistor value you need is R = V/I. Once you know the resistance, you need to determine its power rating. The power dissipated in the resistor is P = V*I, where V is the voltage across it and I is the current through it. Or, you can use P = (I^2)*R. You should choose a power rating at least 1.5 times the power you've calculated so it doesn't run too hot.
2006-11-13 13:27:11 · answer #1 · answered by pack_rat2 3 · 0⤊ 0⤋
it depends on the load...here is the deal, since 6V is just half of the source (12V), therefore, you will need a resistor that has an equal resistance to the load. For example, the load is a light buld of 100 ohms, you need a resistor of 100 ohms to supply 6V to it.
2006-11-14 00:09:14 · answer #2 · answered by alandicho 5 · 1⤊ 0⤋
One resistor doesn't step down a voltage - you need two in series (one after the other) to make a "potential divider".
http://en.wikipedia.org/wiki/Potential_divider
but there are other methods which may be more suitable depending on your application
http://en.wikipedia.org/wiki/DC-DC_converter
2006-11-13 21:23:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Depends on the current. After all, V=IR, emaning
Voltage = Current x Resistance
2006-11-13 21:20:04 · answer #4 · answered by anon 5 · 0⤊ 0⤋
Tell us the current and we'll tell you the answer. Or you can use a 3 Terminal Voltage Regulator.
2006-11-13 23:57:47 · answer #5 · answered by charley128 5 · 1⤊ 0⤋
do the math I over E times R , R is resistence, I is amps, E is voltage
2006-11-13 21:25:20 · answer #6 · answered by Jimmy C 5 · 0⤊ 0⤋
just get r16's
2006-11-14 00:02:09 · answer #7 · answered by Anonymous · 0⤊ 0⤋