The first statement is true, the motor needs a power source to operate.
D.The voltage is low
How do you calculate voltage drop for starting motor current
If you have a simple circuit. For eg: One voltage source and one resistor, then the voltage of the circuit will always remain the same, the current however will decrease following Ohms' Law V=I*R. If we have a current source instead of a voltage source, we are forcing the current to be a certain value so if we increase the resistor value the current will remain the same but the voltage will increase.
Both technicians are right, and both technicians are wrong, because not enough information is present in the question, nor in their statements. Given constant impedance, current should decrease as voltage decreases, while given constant power, current should increase as voltage decreases.
series circuits have 1 pathway they have constant current(Amperes) not constant voltage. Resistance=R+R+R+...
No, There can't Be current without voltage
How do you calculate voltage drop for starting motor current
The voltage drop during starting a load specially motor load is due to absence of back emf and due to high starting current drawn. In case of resistor/heaters, they are also wound like coil so they will also have some inductance and hence during starting, due to absence of back emf, a high current is drawn and hence the voltage drop. While stopping the load current drawn is smaller than the starting current. In the cases of motors and transformers their starting current is about 6 - 8 times the full load current and hence the above phenomenon asked in the question is observed.
If you have a simple circuit. For eg: One voltage source and one resistor, then the voltage of the circuit will always remain the same, the current however will decrease following Ohms' Law V=I*R. If we have a current source instead of a voltage source, we are forcing the current to be a certain value so if we increase the resistor value the current will remain the same but the voltage will increase.
To calculate the starting current of a motor, the run current must be stated. The voltage is only associated with this calculation in as the higher the motor's voltage is the lower the run current is. The start current can be as high as 300 to 600 percent of the run current. Also taken into account, without the specific make and model of the A/C, the run current is hard to guess as there is quite a variety of amperage drawn by these units.
Both technicians are right, and both technicians are wrong, because not enough information is present in the question, nor in their statements. Given constant impedance, current should decrease as voltage decreases, while given constant power, current should increase as voltage decreases.
For example, you can write statements based on:* Ohm's Law: V = IR (voltage = current x resistance) * Power dissipation: P = I squared R
If the ratio of voltage to current is constant, then the circuit is obeying Ohm's Law. If the ratio changes for variations in voltage, then the circuit does not obey Ohm's Law.
normally delta connection wired in 3 phase induction motor. during starting wiring is in Star and after running normal speed changeover to delta .beacause starting time its phase voltage equals less root3 times of line voltage ,line current and phase current equals. in Delta phase voltage and line voltage equals, and phase current equals root3 times line current
One need to know the voltage, the power factor and starting torque, to derive the starting current. The information provided in the question is not sufficient.
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
An example of mass and resistance is voltage, we know that the following is the formula for calculating voltage: V = IR where R is the resistance and I is the current.