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Squirrel Cage Induction Motor Disadvantages:Some of disadvantages or demerits of squirrel cage induction motors are listed below:Main disadvantage of squirrel cage induction motor is that they have poor starting torque and high starting currents. Starting torque will be in the order of 1.5 to 2 times the full load torque and starting current is as high as 5 to 9 times the full load current. In slip ring induction motors, higher starting torque can be attained by providing an external resistance in the rotor circuits during starting of the slip-ring induction motor. This arrangement in slip-ring induction motors also reduces the high inrush currents during starting of induction motor.Squirrel cage induction motors are more sensitive to the supply voltage fluctuations. When the supply voltage is reduced, induction motor draws more current. During voltage surges, increase in voltage saturates themagnetic components of the squirrel cage induction motor.Speed control is not possible in squirrel cage induction motor. This is one of the major diadvantages of squirrel cage induction motors.The total energy loss during starting of squirrel cage motor is more compared to slip ring motors. This point is significant if the application involves frequent starting
What effect will be there on the motor (Induction) output power when a 100kW 50hz motor is connected to a 60hz power supply.
A series-wound commutator motor has the best starting torque because the torque is proportional to the square of the current, and the starting current is set by a current-limiting resistor which is switched out as the motor builds up speed.
If you add one extra bulb and the voltage remains constant, then you have doubled the current drained from the regulator. 12 Volt and One 12 Watt lightbulb drains 1 Ampere Current. 12 Volt and Two 12 Watt light bulbs drains 2 Ampere Current. However: If having a 24 volt powersource and you add two 12 Volt 12 Watt in serial, then you still only drain 1 Ampere Current. NOTE: Wattage and Voltage of bulbs may be different even if the sockets are the same. Lower voltage on the bulb will increase the current drain, if voltage is a lot lower it might cause the circuit delivering voltage to burn out or blow a fuse. It can also quickly burn the bulb, sometimes in a fraction of a second. It will however do little damage to add a bulb with higher voltage than the circuit is designed for. You will then only observe that you do not get the light you might hope for. Total Current/Ampere= Combined Wattage divided by Voltage Total Wattage = Combined Current or Ampere multiplied by Voltage. Regards.
If you add one extra bulb and the voltage remains constant, then you have doubled the current drained from the regulator. 12 Volt and One 12 Watt light bulb drains 1 Ampere Current. 12 Volt and Two 12 Watt light bulbs drains 2 Ampere Current. However: If having a 24 volt power source and you add two 12 Volt 12 Watt in serial, then you still only drain 1 Ampere Current. NOTE: Wattage and Voltage of bulbs may be different even if the sockets are the same. Lower voltage on the bulb will increase the current drain, if voltage is a lot lower it might cause the circuit delivering voltage to burn out or blow a fuse. It can also quickly burn the bulb, sometimes in a fraction of a second. It will however do little damage to add a bulb with higher voltage than the circuit is designed for. You will then only observe that you do not get the light you might hope for. Total Current/Ampere= Combined Wattage divided by Voltage Total Wattage = Combined Current or Ampere multiplied by Voltage. In simpler words: If you double the bulbs, twice the current is drained from the battery
Electron flow, or just current. The more electrons flowing per second, the higher the current.
Squirrel Cage Induction Motor Disadvantages:Some of disadvantages or demerits of squirrel cage induction motors are listed below:Main disadvantage of squirrel cage induction motor is that they have poor starting torque and high starting currents. Starting torque will be in the order of 1.5 to 2 times the full load torque and starting current is as high as 5 to 9 times the full load current. In slip ring induction motors, higher starting torque can be attained by providing an external resistance in the rotor circuits during starting of the slip-ring induction motor. This arrangement in slip-ring induction motors also reduces the high inrush currents during starting of induction motor.Squirrel cage induction motors are more sensitive to the supply voltage fluctuations. When the supply voltage is reduced, induction motor draws more current. During voltage surges, increase in voltage saturates themagnetic components of the squirrel cage induction motor.Speed control is not possible in squirrel cage induction motor. This is one of the major diadvantages of squirrel cage induction motors.The total energy loss during starting of squirrel cage motor is more compared to slip ring motors. This point is significant if the application involves frequent starting
While starting a motor from standstill to its rated speed, the motor has to overcome the inertia and generate enough torque to over come it. In the process the motor takes higher current during the starting. Once started and set in motion the current reduces ti its normal value. Full load current is lower than the starting current normally.
What effect will be there on the motor (Induction) output power when a 100kW 50hz motor is connected to a 60hz power supply.
A series-wound commutator motor has the best starting torque because the torque is proportional to the square of the current, and the starting current is set by a current-limiting resistor which is switched out as the motor builds up speed.
At a steady state (when up and running) Watts = Volts x Amps so you just need to solve the equation 75,000 / 400 = ? However, you add the caveat "starting". If your application involves a motor, for example, the starting current will be higher than the steady state current. One rule of thumb says the starting current may be as high as 6 times the running current. This higher current starts high and then decreases as the motor comes up to speed.
The current relates to the flow of electric charge in an object. Basically the Ampere hour is a measure of how 'long' it will last for. So a higher amount will result in a longer lasting battery because it 'contains' more current.
The camera will draw the amount of current it needs to function. As long as the voltage is within the rating of the camera you are okay.
If you add one extra bulb and the voltage remains constant, then you have doubled the current drained from the regulator. 12 Volt and One 12 Watt lightbulb drains 1 Ampere Current. 12 Volt and Two 12 Watt light bulbs drains 2 Ampere Current. However: If having a 24 volt powersource and you add two 12 Volt 12 Watt in serial, then you still only drain 1 Ampere Current. NOTE: Wattage and Voltage of bulbs may be different even if the sockets are the same. Lower voltage on the bulb will increase the current drain, if voltage is a lot lower it might cause the circuit delivering voltage to burn out or blow a fuse. It can also quickly burn the bulb, sometimes in a fraction of a second. It will however do little damage to add a bulb with higher voltage than the circuit is designed for. You will then only observe that you do not get the light you might hope for. Total Current/Ampere= Combined Wattage divided by Voltage Total Wattage = Combined Current or Ampere multiplied by Voltage. Regards.
If you add one extra bulb and the voltage remains constant, then you have doubled the current drained from the regulator. 12 Volt and One 12 Watt light bulb drains 1 Ampere Current. 12 Volt and Two 12 Watt light bulbs drains 2 Ampere Current. However: If having a 24 volt power source and you add two 12 Volt 12 Watt in serial, then you still only drain 1 Ampere Current. NOTE: Wattage and Voltage of bulbs may be different even if the sockets are the same. Lower voltage on the bulb will increase the current drain, if voltage is a lot lower it might cause the circuit delivering voltage to burn out or blow a fuse. It can also quickly burn the bulb, sometimes in a fraction of a second. It will however do little damage to add a bulb with higher voltage than the circuit is designed for. You will then only observe that you do not get the light you might hope for. Total Current/Ampere= Combined Wattage divided by Voltage Total Wattage = Combined Current or Ampere multiplied by Voltage. In simpler words: If you double the bulbs, twice the current is drained from the battery
mAh stands for milli ampere hour and it is a measurement of how many milli amps of current a battery is capable of producing in one hour. Yes you can substitute a battery with a higher mAh current rating provided you use the same battery voltage.
A current is never concidered negative. The current will always flow from a higher voltage to a lower.For example:12 -> 0V as 12V is a higher potential than 0V. The current is measured in Ampere.0V -> -12V as 0V is a higher potential than -12V. The current is measured in Ampere. Not negative Ampere despite the fact that it floats from the 0V potentialAnswerThe term 'negative', in the sense of 'polarity', doesn't apply to either current or voltage (potential difference), although it does apply to potential -so, the previous answer should read that current 'flows from a higher potential to a lower potential', not 'from a higher voltage to a lower voltage'.However, we do use the terms 'negative' and 'positive' in the sense of 'sense' or 'direction'. So, when we talk about a 'negative voltage', what we mean is a voltage that is acting in the opposite direction to another voltage -for example, if we consider the direction in which a battery's voltage (E) to be acting to be 'positive', then any resulting voltage drop (V) will be acting in the 'negative' direction -i.e. E - V = 0 (i.e. the algebraic sum of the voltages around that circuit is zero).In the same way, we can describe current as being 'positive' or 'negative' in the sense of its direction. For example, if one current (I1) is approaching a junction, and two currents (I2 and I3) are leaving that junction, we can write: I1 - I2 - I3 = 0 (i.e. the algebraic sum of the currents at that junction is zero).Unfortunately, this doesn't specifically answer your question which, unfortunately, is rather confusing and needs to be rephrased.