Since a short circuit is, essentially, a zero impedance connection between nodes, the current in a short circuit is limited only by the ability of the source.
In the case of an ideal voltage source connected to an ideal short circuit, you would have infinite amperes.
Current equals voltage divided by resistance in a DC circuit, or I=V/R (A similar relationship applies to AC.) The "original" circuit has the resistance of the wire plus the resistance of the load. The short circuit only has the resistance of the wire up to the point of the short circuit. In the equation I=V/R, less resistance with the same source voltage results in higher current.
Current is usually very high in a short circuit because the short circuit usually represents a low resistance path for current to flow. By Ohm's Law, current is voltage divided by resistance, and when resistance is very low, current is very high.
The current will be higher and voltage will be zero, during short circuit.
Yes. Voltage becomes zero in a short circuit.
Amps Ohm's law states the current is directly proportional to the applied emf (voltage) and inversely proportional to the resistance of the circuit.
There is a way to use capacitors to increase the voltage in an ac circuit. It increases it by about 50 %. It was used to increase the voltage going to motors. It is seldom used now that it is much easier to run a higher voltage line into a place of business. It does not work for Direct Current.
In a simple circuit, lowering the voltage will not cause the resistance to do anything. Lowering the voltage will, however, cause the current to also lower.This ignores temperature coefficient. If there is substantial power involved, a typical bulb, for instance, will grow cooler and its resistance will decrease when you lower the voltage, but that is usually a small effect.
There is no particular benefit for having a higher open-circuit (or 'no-load') voltage. In fact, an ideal voltage source would have no internal resistance and, therefore, its open-circuit voltage would be identical to its closed-circuit voltage.
No, not directly. The supply voltage has to rise or the resistance has to fall to get over-current. If there was a secondary control voltage that was part of a voltage control circuit for a higher voltage, it is conceivable that a voltage drop in control circuit could cause an over-voltage in the supply. Motors are constant power devices, so this could be true for a motor. If you have a 1hp motor (loaded at 1hp), it will want to draw 1hp of power no matter the supply voltage. If the voltage dips, the motor will require more current to keep it spinning at it's normal speed.
This doesn't make sense, "current" is "amperage" so the higher the voltage the lower the amperage, and the lower the voltage the higher the amperage.
A higher voltage means that a higher current will flow in the same load. It is the current that causes the breaker to trip.
The readings on an ammeter indicate the current being drawn by a load in a circuit. This load is basically a resistance to current flow. The higher the resistance, the lower the current. The supply voltage has a direct effect on current flow. The higher the voltage applied, the higher the current will be. So the readings will vary on the ammeter according to fluctuations in load and or resistance of the circuit and the applied voltage.
Open circuit voltage is the voltage at the electrode before striking an arc (with no current being drawn). The higher the open circuit voltage, the easier it is to strike an arc because of the initial higher voltage pressure.
An electric current relies on several things. First, there must be a continuous connection of conductors. Then there must be a voltage or a potential difference between two parts of a circuit. It is the voltage that causes electrons to move, so generating an electric current. The amount of current depends on both the voltage and the resistance of the circuit. The higher the voltage, the greater the current. The higher the resistance, the lower the current will be.
Amps Ohm's law states the current is directly proportional to the applied emf (voltage) and inversely proportional to the resistance of the circuit.
There is a way to use capacitors to increase the voltage in an ac circuit. It increases it by about 50 %. It was used to increase the voltage going to motors. It is seldom used now that it is much easier to run a higher voltage line into a place of business. It does not work for Direct Current.
In a d.c. circuit, voltage drop is the product of resistance and current through that resistance.
The current (amps) will remain constant, but the voltage will drop.
If you don't change the voltage between the ends of the circuit,then higher resistance in the circuit means lower current (amps).
In a simple circuit, lowering the voltage will not cause the resistance to do anything. Lowering the voltage will, however, cause the current to also lower.This ignores temperature coefficient. If there is substantial power involved, a typical bulb, for instance, will grow cooler and its resistance will decrease when you lower the voltage, but that is usually a small effect.
The resistance of the component on that branch of the circuit, if the resistance is higher less of a proportion of the total current of the circuit will travel through that branch, however, if the resistance is low a higher proportion of the current will travel through that branch of the circuit. The voltage through each branch stays the same.