I saw this and it is completely different from my question Resistance calculation

My question is below. I know that Ohms law is E=IR but I don't know how to use this in a real circuit.

if I have 12 volt battery 24 amps capacity of battery. and I want to start a device which required 0.5 amps and 5 volts. how will i calculate the resistance to reduce the voltage. why amps are necessary here as I know that resistance reduce voltage so it should be required Voltage = E/R so why we always need amps calculation and how will we calculate it in this case?

  • \$\begingroup\$ Voltage is not E/R, but I*R. That's why you need to include amps. \$\endgroup\$
    – gbarry
    May 31, 2015 at 9:19
  • \$\begingroup\$ Point of information : "24 amps" is not a measure of a battery's capacity. "24 amp-hours" or 24Ah would be, which means the battery can last for 48 hours delivering 0.5 amps. \$\endgroup\$
    – user16324
    May 31, 2015 at 11:12

2 Answers 2


I want to start a device which required 0.5 amps and 5 volts.

If all you have is a 12V battery then you need to add a series resistor that drops 7V (12V then becomes 5V) whilst allowing 0.5 amps to the load: -

\$\dfrac{Voltage}{Current} = \dfrac{7}{0.5}\$ = 14 ohms (this is basic ohms law).

If your device always takes 0.5 amps then this will work OK but, if you require 5V to be delivered to your device when the current taken by said device is different then you need a voltage regulator.

Incidentally, whether it's a dropper resistor or a linear regulator the power dissipated in that component will be 7 volts X 0.5 amps = 3.5 watts.

  • \$\begingroup\$ you mean if I need 1amp and 5 volts than i need 7ohms resistance because the current is changed, Good answer and if I need to run 2 devices of 5V which required different amps than I must use a Regulator am I right? \$\endgroup\$
    – Andriopak
    May 31, 2015 at 10:21
  • \$\begingroup\$ Yes, if current is 1A then a 7 ohm resistor is needed BUT, of course this is what a 5V voltage regulator does - it continuously checks the output is 5V and, if it moves a few milli volts higher, it increases the series resistance to compensate. \$\endgroup\$
    – Andy aka
    May 31, 2015 at 10:36
  • \$\begingroup\$ Can you please answer this one, electronics.stackexchange.com/questions/173231/… \$\endgroup\$
    – Andriopak
    May 31, 2015 at 14:19

You probably don't want to use a resistor to reduce the voltage from a battery. This is because the load will also be a resistance so it will influence the resulting voltage which will change with temperature and load etc.

What you need is a voltage regulator. You can use LM7805 which should be suitable for your needs.

They are really simple to use. Only having input, output and ground connections: enter image description here

  • \$\begingroup\$ I don't have any objection on the circuit which you shown me above but I want to know how I can calculate the resistance (not transistor, regulator etc) for this situation, and what are side effects of this? Please \$\endgroup\$
    – Andriopak
    May 31, 2015 at 9:21

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