No one is going to be able to tell you that. You are looking for the measured voltage, so go and measure it. In any case, if you were just looking for the voltage it will depend on the circuit current. You can work it out using ohms law (Voltage = Current * Resistance).
Resistance is measured in ohms.
V = IR Voltage (V, measured in volts) = Current (I measured in amps) times Resistance (R, measured in Ohms) Is your current 3.13? I'm not sure what you wrote there, but just multiply your current times your resistance.
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
V=IR by ohms law. Voltage across the resistor is the product of current flowing and resistance of the conductor at constant temperature.
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
A resistor's resistance is measured in ohms. The higher the resistance the less current will flow with a constant voltage applied across the resistor. In terms of Ohm's Law Voltage = Current x Resistance.
Resistance is measured in ohms.
V = IR Voltage (V, measured in volts) = Current (I measured in amps) times Resistance (R, measured in Ohms) Is your current 3.13? I'm not sure what you wrote there, but just multiply your current times your resistance.
Voltage is not measured in ohms. It is measured in volts.
No, a resistor isn't measured at all. A resistor has a quality called "resistance" - and that value is measured. Resistance is measured in Ohms.
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
V=IR by ohms law. Voltage across the resistor is the product of current flowing and resistance of the conductor at constant temperature.
Don't follow what a '2200watt resistor' is. A resistor spec is measured in ohms. Ohms Law is expressed as: Voltage drop = current x resistance, and the wattage of the resistor is = volts drop x current. You have to decide if your resistor is 2200 ohms, or is taking 2200 watts. These two alternatives will give different results for the current. If it is 2200 watts, at 110 volts, the current is 20 amps. If it is 2200 ohms, at 110 volts, the current will be 50 milliamps. (0.05amps)
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
The ohm. It is how much resistance a component or part of a circuit has to the flow of electrical charge when a voltage is induced across it.
The answer is 6 ohms x 3 amps which is 18 v.
30 ohms x 0.5 amp, that is 15 v.