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.
the current doubles.. explanation:V=IR hence I=V/R which means that when the supply voltage is constant ,current is inversely proportional to resistance.thus the current doubles. practically speaking when the resistance of the load(fan ,bulb,refrigerator,....) is less ,it draws more current from the source so as to balance the voltage across it.i.e; to maintain the voltage across it as constant. This answer is absolutely correct if you assume that the current comes from a pure voltage source ( voltage source with zero internal resistance). At the other extreme you could have a current source (such as a very large voltage source in series with a very large resistor), and then the current is practically independent of changes if the external resistance is changed (because the change represents a relatively minute change in the overall resistance). With appropriate circuitry it is possible to devise a situation where the current is practically independent of the changing resistance.
Nothing. But the current is halved.
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.
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.
it will cause a Short Circuit
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.
If I0 = V/R, then Inew = (2*V)/(.5*R) = (2 / .5) * (V/R) = 4 *V/R = 4 * I0
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.
the current doubles.. explanation:V=IR hence I=V/R which means that when the supply voltage is constant ,current is inversely proportional to resistance.thus the current doubles. practically speaking when the resistance of the load(fan ,bulb,refrigerator,....) is less ,it draws more current from the source so as to balance the voltage across it.i.e; to maintain the voltage across it as constant. This answer is absolutely correct if you assume that the current comes from a pure voltage source ( voltage source with zero internal resistance). At the other extreme you could have a current source (such as a very large voltage source in series with a very large resistor), and then the current is practically independent of changes if the external resistance is changed (because the change represents a relatively minute change in the overall resistance). With appropriate circuitry it is possible to devise a situation where the current is practically independent of the changing resistance.
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.
Nothing. But the current is halved.
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.