It depends on the current going through it.
Ohm's law: Voltage equals current times resistance.
The voltage going into each parallel node will be equal to the voltage of the node before it, as long as the resistance is the same. For example, if you have a current of 5 volts split between two 1k ohm resistors, each 1k ohm will have 5 volts before the resistor drops the voltage.
Its dependent what will by rated power of the device (current).
Yes. You can use a voltage divider. Say, for instance, one 1KOhm resistor in series with a 3KOhm resistor. Connect the 3k resistor to the 48 volts and connect the 1k resistor to ground. The 1k resistor will have 12 volts acress it. These resistors need to be at least 1 watt each as they are going to dissipate 0.576 watts and get warm. Now, if you attempt to pull power from the 1k resistor, note that regulation will be poor because the impedance of the load will go in parallel with the 1k resistor and change its value.
Green, Black, Orange, (gold / silver) Green = 5 Black = 0 Orange = x10^3 50x10^3 = 50,000 or 50k Ohm
In a 12VDC circuit with a 1K load, there will be 12ma of current. (Ohm's law: Volts = Amps * Ohms, so Amps = Volts / Ohms.)
The voltage going into each parallel node will be equal to the voltage of the node before it, as long as the resistance is the same. For example, if you have a current of 5 volts split between two 1k ohm resistors, each 1k ohm will have 5 volts before the resistor drops the voltage.
1,000 ohms and 3,000 ohms in series = total effective resistance is (1,000 + 3,000) = 4,000 ohms.Power dissipated = I2 R.I = sqrt( P / R ) = sqrt (.025/4,000) = sqrt(6.25 x 10-6) = 0.0025 AVoltage drop = I RAcross 1,000 ohms, V = 1,000 I = (1,000 x 0.0025)= 2.5 volts.
the k is the Metric value kilo, or ,1000 so this is a 1000 ohm resistor
It is a passive electrical device with a resistive value of 1000 ohms, used for limiting current or dropping voltage.
Its dependent what will by rated power of the device (current).
dude... using the relation power is the ratio of square of voltage to the resistance P=V^2/R on substitutiing , we get v=(0.000625)^1/2 = 0.125 V
Yes. You can use a voltage divider. Say, for instance, one 1KOhm resistor in series with a 3KOhm resistor. Connect the 3k resistor to the 48 volts and connect the 1k resistor to ground. The 1k resistor will have 12 volts acress it. These resistors need to be at least 1 watt each as they are going to dissipate 0.576 watts and get warm. Now, if you attempt to pull power from the 1k resistor, note that regulation will be poor because the impedance of the load will go in parallel with the 1k resistor and change its value.
Such a circuit would be called a voltage divider.The circuit would consist of two or more resistors in series across a battery or other voltage source. Each resistor would drop a certain amount of voltage (proportional to its resistance), and by considering the voltage drops, the investigator could pick two points in the circuit from which to take (or "pick off") the desired voltage needed for a project. Let's look at just one example.If a 12 volt battery has two 1K ohm resistors in series across it, each resistor will drop 6 volts. By connecting wires from each end of one resistor, the 6 volts can be "picked off" and used to do something else. Certainly there are limitations concerning how much current can be drawn from the circuit (called loading the circuit), as the "diversion" of current around the resistor that is providing the voltage will change the voltage that is being picked off. But for small amounts of current, the voltage divider will work adequately.
Multiply by 1000. 1K ohm = 1000 ohms
Green, Black, Orange, (gold / silver) Green = 5 Black = 0 Orange = x10^3 50x10^3 = 50,000 or 50k Ohm
In a 12VDC circuit with a 1K load, there will be 12ma of current. (Ohm's law: Volts = Amps * Ohms, so Amps = Volts / Ohms.)
It is a passive electrical device with a resistive value of 1000 ohms, used for limiting current or dropping voltage.