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If the excitation of generator changes what happens to the terminal voltage?
By Decreasing the excitation voltage the terminal voltage will decrease and similarly by increasing the excitation voltages the terminal voltage will also increases.
Why the armature current is increasing when the terminal voltage is decreasing?
I did of enginearing pleace help me of you
Why terminal voltage of the self-excited shunt generator lower than that of the separately excited shunt generato?
Some generators are self excited; this means their terminal voltage is fed back to the excitation system to supply power to the rotor of the generator (which makes it into an electromagnet); the more power that is fed back, the stronger the electromagnet becomes, which makes it harder to turn the generator, which causes the generator to push out more power (simplified, really quick version). If there is a fault electrically near the terminal of a self excited generator, the terminal voltage will sage to near zero; this means the voltage supplied to the excitation system will drop by the same percentage (say the terminal voltage is 30% of what it should be, then the maximum supplied voltage to the excitation system drops to 30% of what it normally is, since P = V*I). Since the input power is less, the output of the generator will decrease (current will decrease). The terminal voltage is determined by the impedance between the generator and the fault such that V = I*Z; As I decreases, V will also continue to fall, causing the terminal voltage to sag even more. A non-self excited generator gets its' excitation power from the grid, specifically from a location that is electrically separated from its' terminal voltage. If the terminal voltage sagged to 30% (same fault location as above example), the excitation system voltage may be impacted slightly (say 2%) so the excitation system power is near maximum (98% for this example). Since the excitation system is much farther removed from the terminal voltage, it is not dependent upon it, thus the terminal voltage will not continue to sag as with a self excited system.
What are the factors that influence the speed of the direct-current shunt motor with increasing load?
.The magnitude of the voltage and current of both the armature and shunt field coil. To decrease the speed when the load is increasing then increase the shunt field current while decreasing the armature voltage or current. To increase the speed while the load is increasing then increase the armature current while decreasing the shunt field current. The decreasing and increasing of these currents and voltages can be done by connecting a variable resistor in series or parallel with each of the armature and/or shunt field coil.
How do you reduce kwh by caacitor using?
To reduce kWh by capacitor is when a motor is put in. The terminal voltage is reducing and current is increasing it is connected parallel with the motor.
If the excitation of generator changes what happens to the terminal voltage?
By Decreasing the excitation voltage the terminal voltage will decrease and similarly by increasing the excitation voltages the terminal voltage will also increases.
How do you explain excitation?
by increasing the terminal voltage
Why the armature current is increasing when the terminal voltage is decreasing?
I did of enginearing pleace help me of you
Why The Excitation current is non-sinusoidal when applied voltage is sinusoidal?
excitation voltage is sinusoidal because it is taken from the terminal of alternator but excitation current is non-sinusoidal because it always dc.
What effect if excitation voltage raise on load?
The generator terminal voltage will increase.
What is the formula for excitation voltage?
E=Vt + Ia jXS Where E excitation voltage Vt Terminal voltage Stator Current Ia Xs synchronous Reactance
Is a skydiver increasing or decreasing his speed during the first three seconds?
A skydiver is increasing their speed during the first three seconds of free fall due to gravity pulling them downwards. As the skydiver falls, their speed will continue to increase until they reach terminal velocity.
Why terminal voltage of the self-excited shunt generator lower than that of the separately excited shunt generato?
Some generators are self excited; this means their terminal voltage is fed back to the excitation system to supply power to the rotor of the generator (which makes it into an electromagnet); the more power that is fed back, the stronger the electromagnet becomes, which makes it harder to turn the generator, which causes the generator to push out more power (simplified, really quick version). If there is a fault electrically near the terminal of a self excited generator, the terminal voltage will sage to near zero; this means the voltage supplied to the excitation system will drop by the same percentage (say the terminal voltage is 30% of what it should be, then the maximum supplied voltage to the excitation system drops to 30% of what it normally is, since P = V*I). Since the input power is less, the output of the generator will decrease (current will decrease). The terminal voltage is determined by the impedance between the generator and the fault such that V = I*Z; As I decreases, V will also continue to fall, causing the terminal voltage to sag even more. A non-self excited generator gets its' excitation power from the grid, specifically from a location that is electrically separated from its' terminal voltage. If the terminal voltage sagged to 30% (same fault location as above example), the excitation system voltage may be impacted slightly (say 2%) so the excitation system power is near maximum (98% for this example). Since the excitation system is much farther removed from the terminal voltage, it is not dependent upon it, thus the terminal voltage will not continue to sag as with a self excited system.
What are the factors that influence the speed of the direct-current shunt motor with increasing load?
.The magnitude of the voltage and current of both the armature and shunt field coil. To decrease the speed when the load is increasing then increase the shunt field current while decreasing the armature voltage or current. To increase the speed while the load is increasing then increase the armature current while decreasing the shunt field current. The decreasing and increasing of these currents and voltages can be done by connecting a variable resistor in series or parallel with each of the armature and/or shunt field coil.
Can terminal velocity be changed?
Terminal velocity is determined by several factors including an object's shape, size, and weight, as well as external forces like air resistance. Terminal velocity can be altered by changing these factors, such as by increasing or decreasing an object's weight or by adjusting its shape to reduce air resistance.
What are two ways of increasing the terminal velocity of a falling object?
I'm reluctant to answer because the wording of the question suggests the person asking is looking for answers that meet undefined constraints. One way to increase the terminal velocity of a falling object is to drop it in a vacuum. Another is to drop it in a atmosphere of hydrogen. . 1. increase the mass, without increasing the drag coefficient. 2. Decrease the drag coefficient, without decreasing the mass.
What is the difference between a positive terminal and negative terminal of a battery?
The positive + terminal is slightly larger.
