V=ir
v= (250*10-6) a * (40*103) ohms
v= ( .00025)a * 40000 ohms
v= 10
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.
Output voltage (...of a transformer, for example...) will decrease as it is loaded because of the transformer's internal resistance. As output current increases/load resistance decreases, a larger voltage will be dropped across the internal transformer resistance. This same phenomenon is present in AC and DC systems (such as batteries).
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.
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.
Current flows in loops, voltage drops across elements. With relation to current, what flows in, must flow out, so no, current is not dropped across a resistor, it flows through a resistor and voltage is dropped across the resistor.
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.
As the resistance is reduced across the same voltage, the current increases.
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.
In a d.c. circuit, voltage drop is the product of resistance and current through that resistance.
-- Connect a source of known, small voltage across the ends of the unknown resistance. -- Measure the resulting current through the unknown resistance. -- Divide (small known voltage)/(measured current). The quotient is the formerly unknown 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.