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it is better to use zener diodes or transformators in order to drop 12 vdc to 2 vdc. If you do it with resistor you will waste you power.
Voltage formula for resistors in series V1+V2+........+Vn=Vtotal, so 2Vdc+i.R=12Vdc. Determine the current for you supplier and calculate the value of resistance.
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How do I design a voltage divider circuit that has 2 VDC 5 VDC and 8 VDC available with a 10 VDC source?
Without knowing permissible current draw by the divider or its maximum power dissipation the actual resistor values cannot be determined, but the ratios of the resistor values can be determined from the required voltage drops.The divider will be composed of 4 resistors starting at the 10VDC rail:2VDC drop, ratio = 2V/10V = 0.23VDC drop, ratio = 3V/10V = 0.33VDC drop, ratio = 3V/10V = 0.32VDC drop, ratio = 2V/10V = 0.2Therefore you will need 2 resistors (R1 & R4) that are 0.2 * the total resistance of the voltage divider and 2 resistors (R2 & R3) that are 0.3 * the total resistance of the voltage divider.But as stated at the beginning you can get no further without additional requirements being specified.
Ripple factor of half wave rectifier of L-section with derivation?
ripplefactor(R)=vrms(Ac)\vdc=(v^2rms-v^2dc)^2vdc=vmax/22/7vrms=vmax
A meter should indicate how many volts when connected to a 100 ohm resistor that has 6 amps of current passing through it?
600 VDC.
If a 1.5 V battery is connected to a 330 Ohms resistor Calculate the current in the circuit?
Ohm's laws says it will be 1.5 vdc divided by 330 ohms.
How much ohms resistor is needed to drop 12 volts to 5 volts?
The size of the resistor will depend on the load. Let's look at this a bit to see if we can make sense of it. You want to drop the applied voltage to a device from 12 volts AC to 11 volts AC. That means you want to drop 1/12th of the applied voltage (which is 1 volt) across the resistor so that the remaining 11/12ths of the applied voltage (which is 11 volts) will appear across the load. The only way this is possible is if the resistor has 1/11th of the resistance of the load. Here's some simple math. If you have an 11 ohm load and a 1 ohm resistor in series, you'll have 12 ohms total resistance ('cause they add). If 12 volts is applied, the 1 ohm resistor will drop 1 volt, and the 11 ohm load will drop the other 11 volts. A ratio is set up here in this example, and each ohm of resistance will drop a volt (will "feel" a volt) across it. See how that works? If the resistance of the load is 22 ohms and the resistance of the (series) resistor is 2 ohms, each ohm of resistance will drop 1/2 volt, or, if you prefer, each 2 ohms of resistance will drop 1 volt. The same thing will result, and the load will drop 11 volts and the series resistance will drop 1 volt. That's the math, but that's the way things work. You'll need to know something about the load to select a series resistance to drop 1/12th of the applied voltage (which is 1 volt) so your load can have the 11 volts you want it to have. There is one more bit of news, and it isn't good. If your load is a "dynamic" one, that is, if its resistance changes (it uses more or less power over the time that it is "on"), then a simple series resistor won't allow you to provide a constant 11 volts to that load. What is happening is that the effective resistance of the load in changing over time, and your resistor can't "keep up" with the changes. (The resistor, in point of fact, can't change its resistance at all.) You've got your work cut out for you figuring this one out.
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