If the resistance of the load is kept more-or-less constant, then the current also becomes larger. On the other hand, if the power of the load is kept more-or-less constant, then the current becomes smaller.
That depends on whether the circuit that the voltage is applied across is ohmic (linear) or nonohmic (nonlinear). If the circuit is ohmic then it follows ohm's law and the current is directly proportional to the voltage. If the circuit is nonohmic many different things can happen depending on the details of the circuit (e.g. current goes up when voltage goes up, current goes.down when voltage goes up, current holds constant independent of voltage).
the higher the voltage the more energy is released per charge
In a normal circuit, with a resistor for example, the voltage and current are in proportion, which is known an Ohm's law.
The electric current also increases(provided resistance remains constant) , as current is directionally proportional to voltage.
V=IR(ohm's law)
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.
Crystal diode for a p-type semiconductor and n-type semiconductor formation of the pn junction, in its interface on both sides of a space-charge layer, and has a self-built electric field. When there is no applied voltage, as pn junction on both sides of carrier concentration caused by the proliferation of poor self-built electric current and drift arising from the current equivalent and the balance of power in the state. When the outside world a positive bias voltage, electric and outside the field of mutual self-suppression role of the Consumers carrier increase from the current spread of the forward current. When the outside world a reverse bias voltage, external electric field and to further strengthen self-built electric field, in a certain form of reverse voltage and reverse bias voltage value unrelated to reverse saturated current I0. When the reverse voltage applied to a certain high level, pn junction in the space charge of the electric field strength to achieve the critical values of the double-carrier process, a large amount of electronic hole right, had a great numerical breakdown of the reverse current, Breakdown phenomenon known as diodes.
Very unlikely. Its 3 phase for a reason, it needs a large voltage/current to power it, single phase won't provide that.
The operation of the television, as all electronics, involves electric currents. If your TV uses a CRT display as it operates it happens to build a large static electric field on its face, as a side effect of that operation.
resistors - and they have to be physically large enought to dissipate the heat of the voltage x current you will drop out.
They get electricuted.
Resistor is a component of an electric circuit that resist the flow of direct or alternating electric current . It can also limit or divide the current, reduce the voltage, protect an electric circuit, or provide large amounts of heat or light.
The combination of supply voltage and load. For any given supply voltage, the greater the load (i.e. the lower its resistance), the greater the resulting current.
Over voltage, severe under frequency, or add too large of a secondary burden (the last is what typically happens).
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.
ICE current leads the voltage by 90 degrees.
Crystal diode for a p-type semiconductor and n-type semiconductor formation of the pn junction, in its interface on both sides of a space-charge layer, and has a self-built electric field. When there is no applied voltage, as pn junction on both sides of carrier concentration caused by the proliferation of poor self-built electric current and drift arising from the current equivalent and the balance of power in the state. When the outside world a positive bias voltage, electric and outside the field of mutual self-suppression role of the Consumers carrier increase from the current spread of the forward current. When the outside world a reverse bias voltage, external electric field and to further strengthen self-built electric field, in a certain form of reverse voltage and reverse bias voltage value unrelated to reverse saturated current I0. When the reverse voltage applied to a certain high level, pn junction in the space charge of the electric field strength to achieve the critical values of the double-carrier process, a large amount of electronic hole right, had a great numerical breakdown of the reverse current, Breakdown phenomenon known as diodes.
the lights would grow dimmer if the large appliance draws so much current that the resistance of the main electrical service conductors to that current causes a service voltage drop. less voltage means less light. one reason the lights might grow brighter when the large appliance comes on is if the rest of the household electric load (other than the large appliance)is mostly connected to just one of the service conductors, with the large appliance on the other service conductor and the service neutral is marginal or undersized. the neutral carries the unbalanced current of the two service conductors so, without the large appliance on line, the neutral is carrying a large current and the resistance of the undersized neutral to that large current is causing a voltage drop all the time, so lights are dim all the time. when the large appliance comes on line, the unbalance is reduced, the neutral carries less current, the voltage drop reduces and the lights brighten.
If the resistance is large enough, then there might not be enough voltage difference to allow much current. Since, Voltage = Current * Resistance, if resistance goes really large, and your voltage doesn't change, your current must decrease. An open circuit is where you do not have any current flowing, so whether no current verses very little current is the same is up to you.
A conductor of electricity is one that will carry electric current, without generating undue heat or causing a large voltage drop along it. Copper is a good example.
The starter solenoid receives a large electric current from the car battery and a small electric current from the ignition switch. When the ignition switch is turned on, a small electric current is sent to the starter solenoid. This causes the starter solenoid to close a pair of heavy contacts, thus relaying a large electric current to the starter motor, which in turn sets the engine in motion.
It is not a shunt with zero resistance. It is very small, but it is not zero. The large current develops a small voltage across the small resistance. Measuring that small voltage gives you a proportional measurement of the current.