What will be effect on motor current when an induction motor is supplied with higher voltage than rated voltage?

The current will be higher than the rated full load amps that is stamped on the name plate.

The effect of low voltage on induction motors raises the current that the motor will draw.

RPS is only the voltage& power controlled device. it can only used for set the input for our wish A device which can change its output according to the voltage supplied to it is called a voltage controlled device.ex. a voltage controlled current source,or a field effect transistor. In a voltage controlled current source the output current changes as the voltage supplied to it changes.

Capacitors resist a change in voltage. It takes current to effect a voltage change, resulting in the current "leading" the voltage. Similarly, inductors resist a change in current. It takes voltage to effect a current change, resulting in the current "lagging" the voltage.

You apply a voltage across a load and the result is that a current flows through the load. So you must have the voltage present, the cause, before current flow, the effect. Think of voltage as pressure and current as flow.

Resistance = Voltage ----------- Current So if both the resistance and voltage have halved, the current has doubled

If the current is held constant, the voltage will decrease.

Ohm's Law: Current is voltage divided by resistance. Doubling both the voltage and the resistance will not change the current.

The r.m.s. value of an alternating current or voltage is the value of direct current or voltage which produces the same heating effect.

Ohm's Law states that Voltage = Resistance (Ohms) * Current (Ampere). So when you increase voltage, you increase current.

When a load current flows, as well as flowing through the load, it also flows, internally, through the voltage source. A load current, therefore, causes an internal voltage drop which, by Kirchhoff's Voltage Law, opposes, the electromotive force of the source. The larger the load current, the greater the internal voltage drop. So, the terminal voltage of a voltage source will decrease as the load current increases. By definition, the e.m.f. of the source is equal to the sum of all the voltage drops (including the internal voltage drop) around any closed loop supplied by that source.

If resistance is halved while voltage remains constant, the current will double.

current is unchanged or remains the same

The Hall effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor and a magnetic field perpendicular to the current. It was discovered by Edwin Hall in 1879

It depends on how the capacitor is connected and whether the supply voltage is a.c. or d.c. Assuming you are talking about a power-factor improvement capacitor (connected in parallel with an inductive load, supplied with a.c.), then the supply current will reduce.

That depends on the circuit - but note that in almost all real circuits the current is the dependent variable - you control the voltage and the current sets itself.

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.

Since current is, by Ohm's Law, equal to voltage divided by resistance, then doubling both will result in no net effect to the current.

There will be no effect on the voltage. That is the effective voltage will be only 12 volt. But there will be increase of current.

Ohm's Law says that Voltage = Current x Resistance (Load). Therefore Current = Voltage / Resistance and as resistance decreases current increases and as resistance increases current decreases.

Using Ohms Law: V = I x R, where V (Voltage), I (Current), and R (Resistance). re-arranging: V/R = I Therefore if you double both the Voltage and the Resistance, the current remains unchanged. Current = Voltage / Resistance. If both resistance and voltage double the current remains the same.

Ohm's law: current is voltage divided by resistance, so doubling the voltage while holding the resistance constant will double the current. Not asked, but answered for completeness sake; since power is voltage times current, doubling the voltage into a constant resistance will quadruple the power.

The higher the load current the higher the ripple voltage across the filter capacitors. It is directly proportional.

The effect of an RL circuit in half wave rectifier is that the voltage output wave forms for current and voltage will be modified .

There is no such thing as 'alternative voltage'! The term you want is 'alternating voltage'. The average value of an a.c. voltage or current over a complete cycle is zero, so it is not very useful to think of a.c. in terms of average values. Instead, we look at its 'heating effect' of the a.c. current, and compare it with the heating effect of a continuous, d.c. current. If a current of say, 10 A d.c. produces a particular heating effect, then an a.c. current of 10 A (rms) will result in exactly the same heating effect. This is why 'rms' values are also called 'effective' values, as it has exactly the same effect as a d.c. current of the same value.