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By Ohm's Law, current is voltage divided by resistance, so if you double both the voltage and the resistance, the current would remain the same.
Their relationship is only dependent on the voltage lost across that resistor; voltage equals resistance times current, so increasing the current for a given voltage will require a decrease in the resistance, and vice versa.
Voltage across a resistance = (resistance) x (current through the resistance) =4 x 1.4 = 5.6If the ' 1.4 ' is Amperes of current, then the required voltage is 5.6 volts.
Compute the open load voltage of the current source across its shunt resistance.This voltage becomes the voltage source's voltage.Move the current source's shunt resistance to the voltage source's series resistance.Insert the new voltage source into the original circuit in place of the current source.
You measure current by inserting an amp meter in series with the circuit or using a clamp-on meter to measure current by induction. You measure voltage with a volt meter across the supply. You measure resistance in ohms across the resistance. Luckily all these functions are in the same meter in most cases.
IR drop across a resistance is voltage. The letter I means current, and the letter R means resistance. Current times resistance, by Ohm's law is voltage.
Ohm's law states that the voltage across a resistor is the product of the current times the Resistance or V=I x R (I times R). V is Voltage, R is Resistance, and I is Current or Amperage. So if the Voltage is doubled and Resistance stays the same, the Current will be doubled.
The reason an AC voltage applied across a load resistance produces alternating current is because when you have AC voltage you have to have AC current. If DC voltage is applied, DC current is produced.
By Ohm's Law, current is voltage divided by resistance, so if you double both the voltage and the resistance, the current would remain the same.
As the resistance is reduced across the same voltage, the current increases.
In a d.c. circuit, voltage drop is the product of resistance and current through that resistance.
The power dissipated across a resistor, or any device for that matter, is watts, or voltage times current. If you don't know one of voltage or current, you can calculate it from Ohm's law: voltage equals resistance times current. So; if you know voltage and current, power is voltage times current; if you know voltage and resistance, watts is voltage squared divided by resistance; and if you know current and resistance, watts is current squared times resistance.
Their relationship is only dependent on the voltage lost across that resistor; voltage equals resistance times current, so increasing the current for a given voltage will require a decrease in the resistance, and vice versa.
Voltage across a resistance = (resistance) x (current through the resistance) =4 x 1.4 = 5.6If the ' 1.4 ' is Amperes of current, then the required voltage is 5.6 volts.
Compute the open load voltage of the current source across its shunt resistance.This voltage becomes the voltage source's voltage.Move the current source's shunt resistance to the voltage source's series resistance.Insert the new voltage source into the original circuit in place of the current source.
The amount of current that will pass through a resistance is dependant upon the voltage applied across the resistance. Voltage devided by resistance equals current. This is Ohm's Law.
You measure current by inserting an amp meter in series with the circuit or using a clamp-on meter to measure current by induction. You measure voltage with a volt meter across the supply. You measure resistance in ohms across the resistance. Luckily all these functions are in the same meter in most cases.