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ohms law

Akans

how did i use ohms low to choose a resistor for my circuit when the current of the voltage is unknown. like i have a 10 volt resistor and i need 5volt out from the 10volt what will i do to get the current so that i can use it to divide the voltage to get my resistor for the circuit as ohms low states

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  • 0 in reply to JWx

    what i am trying to say is how to choose a resistor for a led all tutorial i have watch seams they get their current from datasheet and they use ohms low to work out and get the suitable voltage for the led and i have been founding it difficult choose a resistor for my circuit because using ohms low is voltage divide by current and i am using batter so what should i do

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  • 0 in reply to Akans

    for driving a LED you need three values: power supply voltage, diode forward voltage and diode current.

    Diode current can usually be in the range of about 2-20mA

    Diode voltage depend on the color and is standard for that color - for example, yellow LED has forward voltage of about 2.1V

    then, all the remaining voltage should be across the resistor - so (for example, using 9V battery and 5mA)

    9V - 2.1V (diode forward voltage) = 6.9V (voltage drop across resistor)

    6.9V/5mA = 1380 Ohm

    this is not a standard value, but you can use - for example - 1.3kOhm (1300Ohm)

  • 0 in reply to JWx

    what about if i want to drive a 12volt motor and i have a voltage supply to be 18v what should i do

  • 0 in reply to Akans

    install a voltage regulator - switching one if possible to limit power loss.

    Motor has a different characteristics - it consumes different current depending on the load. You can probably limit the current using resistor but it would be wasteful - if you calculate the resistor for idle current, any load increase would increase voltage across resistor, decreasing voltage across motor and limiting power. And if you calculate resistor for loaded motor, you will get overvoltage when motor is idle.

    Anyway - if you can tell what current   will be drawn during operation (for example - steady load), the method is the same - all the excess voltage should be across resistor.

    for example - motor consumes 2A during operation, and voltage difference is 18V-12V = 6V

    6V/2A = 3 Ohm

    and power dissipated would be 6V*2A = 12W

    and of course - if motor is stalled, most of the input voltage would be across the resistor, resulting in 18V * 18V/3Ohm = 108W of power

  • 0 in reply to Akans

    What does this motor do? What is the current or power rating for the motor? Most 12V motors can take 18V for a short time if you just need to test something, but it may depend on how heavily the motor is loaded.